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Some very interesting tentative conclusions can be drawn from this experiment First, compare the average pull-off force of the eight 3 32-inch prototypes with the average pull-off force of the eight 1 8-inch prototypes (these are the averages of the eight runs on the left face and right face of the cube in Fig S1-1, respectively), or 141 1345 065 Thus, increasing the wall thickness from 3 32 to 1 8-inch increases the average pull-off force by 065 pounds Next, to measure the effect of increasing the cure time, compare the average of the eight runs in the back face of the cube (where time 24 hours) with the average of the eight runs in the front face (where time 1 hour), or 14275 13275 1.

The effect of increasing the cure time from 1 to 24 hours is to increase the average pull-off force by 1 pound; that is, cure time apparently has an effect that is larger than the effect of increasing the wall.

The plaintext message CAT is encoded as 000 001 010. Suppose the key K = 110 100 001 is used for encryption. Then the ciphertext is given by C = 110 101 111 which corresponds to EGN. Now, suppose Trudy intercepts C and she guesses the putative key K = 010 110 010. Using K , Trudy computes the putative plaintext

14.8 (14.6, 15.0)

15.1 (14.9, 15.3)

13.0 (12.5, 13.5)

c 69 K = 110 101 111 @ 010 110 010 = 100 011 101

3 32 1 8

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Camera and Shooting Functions Shooting Menu AF Frame Digital Zoom AF-point Zoom Servo AF Continuous AF AF-assist beam MF-point Zoom Safety MF Wind Filter Flash Control i-Contrast Spot AE Point Safety Shift Review Review Info Blink Detection Custom Display Reverse Display IS mode Converter Date/Time Stamp Record RAW +JPEG Set Shortcut button Save Settings Set-up Menu Mute Volume Sound Options Hints and Tips LCD Brightness Start-up Image Format File Numbering Create Folder Lens Retract ..

13.1 (12.9, 13.3)

which corresponds to the message DOG. Based on the ciphertext and the putative plaintext, Trudy has no way to judge whether the message DOG is any more likely than the message CAT, or any other three letter message that car1 be spelled from the eight letters in Table 1.8. That is, decrypting C with any one of the possible 83 = 512 keys gives one of the 512 possible plaintext messages and the ciphertext itself gives no hint as to which of these is correct.

14.1 (13.9, 14.3)

Wall thickness (in.) 12.9 (12.6, 13.2) 13.6 (13.4, 13.8)

The one-time in the one-time pad is crucial. If a key K is used more than once, then the one-time pad (which is, technically, no longer a one-time pad) is subject to an attack. Different messages encrypted with the same key are said to be in depth. Suppose that plaintext messages Po and PI are in depth, that is, both are encrypted with a one-time pad using the same key I(, yielding ciphertexts Co and C1, respectively. Then if Trudy obtains both ciphertexts, she can compute

Time Temp. Avg. Force 1 1 24 24 h 70 F h 100 F h 70 F h 100 F 13.25 13.30 13.60 14.95 15.30 Temp. = 100 F 14.83 14.37 Pounds 13.90 Temp. = 100 F 13.43 12.97 12.50 1h Time 24 h Temp. = 70 F

EXAMPLE S5-3 Let X be a continuous random variable with probability distribution fX 1x2 x , 8 0 x 4

Find the probability distribution of Y h(X ) 2X 4. Note that y h(x) 2x 4 is an increasing function of x. The inverse solution is x u( y) (y 4) 2, and from this we nd the Jacobian to be J u 1 y2 dx dy 1 2. Therefore, from S5-3 the probability distribution of Y is fY 1 y2 1y 42 2 1 a b 2 y 32 4 , 4 y 12

that is, Trudy can obtain the XOR of the two plaintexts It might then be possible for Trudy to peel apart these two messages, depending on the properties of the plaintext The fundamental issue here is that the attacker can, in effect, use one of the messages as a check on any guess for the other message (or the key) Consequently, the ciphertext now provides information about the underlying plaintext and the security can no longer be assured The problem only gets worse (or, from Trudy s perspective, better) the more the one-time pad is reused An obvious practical problem with the one-time pad is that a key having the same length as the plaintext must be securely transmitted to the recipient, and this key can only be used once.

We now consider the case where X1 and X2 are continuous random variables and we wish to nd the joint probability distribution of Y1 h1(X1, X2) and Y2 h2(X1, X2) where the transformation is one to one. The application of this will typically be in nding the probability distribution of Y1 h1(X1, X2), analogous to the discrete case discussed above. We will need the following result.

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