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| | Cryptanalysis of MultiSwap (Site not responding. Last check: 2007-11-07) |
 | | For example, to force the value multiplied by k6 to be w, simply query the encryption oracle with plaintext (0,w-k5). |
 | | One can repeat this test for 16 right input pairs (w1,2*w1)...(w16,2*w16) chosen uniformly at random, and the probability of a given k6 value surviving all 16 tests is roughly (1/4)^16 = 2^-32, so we expect about one value of k6 to survive. |
 | | Thus with 2^22.5 known plaintexts, we expect that the 2^22.5 inputs to the second round will contain about 2^13 pairs, enough to recover k6,...,k10. |
| www.cs.sunysb.edu /~rtjohnso/multiswap (1884 words) |
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