tag:blogger.com,1999:blog-4115025577315673827.post5606554537205953497..comments2020-05-20T14:21:54.596+05:30Comments on CSE Blog - quant, math, computer science puzzles: Spy Control Problem - Peter WinklerPratik Poddarhttp://www.blogger.com/profile/11577606981573330954noreply@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 /> Anonymoushttps://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??nikhilhttps://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