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222 CHAPTER 11. KNAPSACK CIPHERS
77 = (3) + 6 + 13 + 55. Yes to 3.
77 = 3 + 6 + 13 + 55. Done. Found a total that works.
(3) Same weights. Demand is for 81.
81 = (81). Yes to 55.
81 = (26) + 55. No to 27. Yes to 13.
81 = (13) + 13 + 55. Yes to 6.
81 = (7) + 6 + 13 + 55. Yes to 3.
81 = (4) + 3 + 6 + 13 + 55. Yes to 2.
81 = (2) + 2 + 3 + 6 + 13 + 55.
Done. We didn’t find a total that was correct, and since we can only use
each weight once, this means there is no solution.
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄
In general, when the numbers or weights are super-increasing, the greedy
method will find a solution when there is one, and will indicate that no solution
is possible when it is impossible.
Example: Weights 3, 7, 12, 24, 49, 104, and 215. Which of the totals 298,
421, 358 and 311 can be found 3 ⋄
Thus, if the weights are super-increasing the knapsack problem is very easy.
If they are not, it can be very very difficult. Next we transform the easy knapsack
problem into (apparently) a hard one.
Example: Consider the five weights W 1 = 3, W 2 = 7, W 3 = 12, W 4 = 24
and W 5 = 49. These are in super-increasing order and so form the weights of
an easy knapsack problem. For example, can these weights produce the total of
80 Yes: 80 = 49 + 24 + 7.
Now let e = 39 and P = 101. e is our “enciphering” multiplier. If we
multiply the W ’s by e modulo P we obtain new weights, U 1 through U 5 :
W 1 × e = 3 × 39 ≡ 16 ≡ U 1 (mod 101)
W 2 × e = 7 × 39 ≡ 71 ≡ U 2 (mod 101)
W 3 × e = 12 × 39 ≡ 64 ≡ U 3 (mod 101)
W 4 × e = 24 × 39 ≡ 27 ≡ U 4 (mod 101)
W 5 × e = 49 × 39 ≡ 93 ≡ U 5 (mod 101)
The weights 16, 71, 64, 27 and 93 are not in super-increasing order! Can the
total 90 (= (80 × 39)%101) be produced from them 4 We have succeeded in
3 Only 298 and 358 are possible.
4 Yes, but only modulo 101: 90 = (93 + 27 + 71)%101.