tag:blogger.com,1999:blog-4115025577315673827.post5606554537205953497..comments2019-10-21T14:10:54.375+05:30Comments on CSE Blog - quant, math, computer science puzzles: Spy Control Problem - Peter WinklerUnknownnoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4115025577315673827.post-32473251685451893372013-10-07T16:12:40.812+05:302013-10-07T16:12:40.812+05:30Is it always possible? If my number is 11110100000...Is it always possible? If my number is 1111010000010100 (sum of the positions is 70), mow I want to transmit 0 (I can't do it by flipping only one bit).<br /> Shivani Agrawalhttps://www.blogger.com/profile/05077453657636718252noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-63058490750912573672013-02-03T01:05:30.706+05:302013-02-03T01:05:30.706+05:30Encryption Algorithm:
To be sent message x (a 4 b...Encryption Algorithm: <br />To be sent message x (a 4 bit number)<br />Broadcast message (a 16 bit number)<br />Cipher Text = Broadcast message xor (A 16 bit number with 0s at all places but 1 at one place) such that (Sum of all the positions where cipher text is 1 mod 16 = x)<br /><br />Decryption Algorithm:<br />Take positions of 1 in the message broadcast. Take sum of all these numbers. Do mod 16 on the number. Represent the result in binary format. You get the message back.<br /> <br />Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-49270881398683728342012-12-26T06:54:26.537+05:302012-12-26T06:54:26.537+05:30but how would the handler know that the third bit ...but how would the handler know that the third bit has been flipped when you want to send 1111??Nikhil Simha Rhttps://www.blogger.com/profile/09880727972082313533noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-82280309415338760262012-12-19T08:53:21.581+05:302012-12-19T08:53:21.581+05:30I checked the solution. It is like a secret key.th...I checked the solution. It is like a secret key.the receiver and sender share a code which is ex-ored with the missing place holder's binary value. But my brute force approach would be check the 15 digit code and change the one bit which u want to send in binary. Like if u want to send 0011 then flip the value of 3rd bit and send it across.I agree that this is not a secure method and it can be tapped. But just a thought. The approach he used reminds me of CDMA.INNER VOICEhttps://www.blogger.com/profile/15591908437934009602noreply@blogger.com