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aligh1979
Posts: 121
Posted 18:05 Oct 16, 2011 |

following slide 25 of Lecture 5 , slide 26 describing that a change in position 8 will cause 4 other positions to change .

I know that because of the SBox , every 2 bits are encrypted (substituted  ) together , but it is a XOR operation , and a XOR operation is one by one bit and only get applied to one bit . so I do not still have a correct understanding of why changing one bit (here bit 8) cause other bits to change. even if we consider that 8 i a letter and contains 4 or 8 bits , then changing that letter does not cause the other letter (2 by 2 in a SBox) to change

Redalb
Posts: 22
Posted 23:39 Oct 16, 2011 |

I agree,  2 bits are changed on the right. Those 2 bits are independent of the other 2 bits to the left and there isnt an operation that use all 4 bits. It's confusing, it seems like it would only change 2 bits. We'll have to get some clarification on this.

hpguo
Posts: 139
Posted 10:04 Oct 17, 2011 |

After XOR, only 1 bit is changed, but after S-Box, 2 bits will be changed.  S-Box is substitution, not XOR.