A quantum computer at IBM's Thomas J. Watson Research Center in New York.The question is if all four quantum computers are identical and how the quantum logic is handled. |
Now a different example: The number 187 which is the product of 11 and 17. In the middle lies the number 14 and 14 + 3 = 17 and 14 - 3 = 11.
Your strategy is calculate the sum of the two prime numbers, divided by 2, because when you know the sum you can calculate the two prime numbers. In this case this is 28/2 =14.
Consider the number 187 which we call n. The next highest square = 196, which is equal to 14^2, and 14 is the value in the middle.
Next we try 11 and 31. n = 11*31 = 341. (pm1+pm2)/2 =21. And 21+10 =31 and 21-10 =11 The next highest square = 361, which is equal to 19^2 but! 19 is not the value in the middle. So finally, this strategy is not correct. Let us follow a slightly different strategy.
X xn 1 2 4 8 16 32 64 128 69 138 89 178 169 151 115 43 86 172 157 127 67 134 81 162 137 87 174 161 135 83 166 145 103 19 38 76 152 117 47 94 a 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 amin xn 1 2 4 8 16 32 64 128 69 138 89 178 169 151 115 43 86 172 157 127 67 134 81 162 137 87 174 161 135 83 166 145 103 19 38 76 152 117 47 94 a 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 X xn 1 2 4 8 16 a 80 81 82 83 84 amax Table 1The above figure shows a repetition of two lines. The bottom line is the value a going from 0 to 84. The top line is the value xn, the function 2^a mod 187. At position a = 39, xn = 94. At position a = 40, xn = 1. The reason is that 2 * 94 - 187 = 1
Step 5 in this strategy is the most difficult. This involves to calculate 2^amax mod 187 to calculate the stop sign 2.
Step 6 involves the same strategy in reverse order. Here your target is the stop sign, while using the function 2^a mod n, for a very small numbers of a to calculate amin. In fact, this is the most time consuming part.
pm1 pm2 n (n+1)/2 sqr(n) amax stopsign amin period (pm1+pm2)/2 11 17 187 94 13 81 2 1 80 14 11 31 341 171 18 153 8 3 150 21 79 139 10823 5412 104 5308 16 4 5304 108 3347843 3674639 12302114453677 6151057226839 3507437 3804 3511241 103 msec 15538213 16860433 261980999226229 130990499613115 16185827 13496 16199323 102 msec 36746137 85478221 3140994419382277 1570497209691139 56044575 5067604 61112179 4800 msec Table 2The above table 2 shows results of factorization of 6 combinations of pm1 and pm2.
To calculate amin more or less goes in reverse order. In this case the stop sign is known, which is the value of the function f(a,x,n).
Now you have to calculate the smallest value of a which as result shows the stop sign. You start with a=1 etc.
However in this case you can use the multiprocessor capability of your PC. Suppose there are 4 multiprocessors or PPs. In this case PP1 can calculate: a from 1 to 1000, PP2 a from 1001 to 2000, PP3 a from 2001 to 3000 and PP4 a from 3001 to 4000.
What is the issue.
Generally speaking I am the only one who uses my PC, as such my PC does not require a password.
However my PC is connected to the outside world, via the Internet, and I want to maintain that capability. However under certain strict rules.
Via the Internet I want to send and receive e-mails, but here also under strict rules.
The first rule is that in general all communication goes via text messages.
The second rule is that in general no one is allowed to make any modification on my PC. That means in short: No cookies.
The exceptions are trusted links. Those trusted links are those required to update the installed software.
This raises the following question: Is there any reason why communication should be encrypted? What is it that I want to hide? Who wants to send me secret information?
That does not mean that certain organizations, for their internal operations, use secret information, but in general most communication is not. The issue is much more how do these organizations, that require secrecy, handle this secrecy internally.
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