How many times was a 8 digit binary number circulary shifted? [closed]

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Given an 8 digit binary number for example: $ 00000111 $, and the same number circularly shifted $k$ times, where $k < 8$, $ 11000001 $ can i calculate k without curcularly shifting the original number until it matches the second one?

computer-science binary

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edited Jul 15 at 18:05

asked Jul 15 at 17:42

Kratos

12

closed as off-topic by amWhy, Xander Henderson, Leucippus, Parcly Taxel, user190080 Jul 16 at 10:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus, Parcly Taxel, user190080

If this question can be reworded to fit the rules in the help center, please edit the question.

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up vote
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Given an 8 digit binary number for example: $ 00000111 $, and the same number circularly shifted $k$ times, where $k < 8$, $ 11000001 $ can i calculate k without curcularly shifting the original number until it matches the second one?

computer-science binary

share|cite|improve this question

edited Jul 15 at 18:05

asked Jul 15 at 17:42

Kratos

12

closed as off-topic by amWhy, Xander Henderson, Leucippus, Parcly Taxel, user190080 Jul 16 at 10:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus, Parcly Taxel, user190080

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

Given an 8 digit binary number for example: $ 00000111 $, and the same number circularly shifted $k$ times, where $k < 8$, $ 11000001 $ can i calculate k without curcularly shifting the original number until it matches the second one?

computer-science binary

share|cite|improve this question

edited Jul 15 at 18:05

asked Jul 15 at 17:42

Kratos

12

Given an 8 digit binary number for example: $ 00000111 $, and the same number circularly shifted $k$ times, where $k < 8$, $ 11000001 $ can i calculate k without curcularly shifting the original number until it matches the second one?

computer-science binary

share|cite|improve this question

edited Jul 15 at 18:05

asked Jul 15 at 17:42

Kratos

12

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 18:05

edited Jul 15 at 18:05

edited Jul 15 at 18:05

asked Jul 15 at 17:42

Kratos

12

asked Jul 15 at 17:42

Kratos

12

asked Jul 15 at 17:42

Kratos

12

12

closed as off-topic by amWhy, Xander Henderson, Leucippus, Parcly Taxel, user190080 Jul 16 at 10:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus, Parcly Taxel, user190080

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Xander Henderson, Leucippus, Parcly Taxel, user190080 Jul 16 at 10:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus, Parcly Taxel, user190080

If this question can be reworded to fit the rules in the help center, please edit the question.

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add a comment | 

2 Answers
2

active

oldest

votes

up vote
0
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accepted

You can't calculate $k$ at all, even by circularly shifting. Suppose you shifted it $j$ times (for some $jin0,1,2,3,4,5,6,7$ to get the original number. Then for any nonnegative integer $n,$ shifting the original number $8n-j$ times looks exactly the same as it would for any other such $n.$

The best you can do is calculate $k$ modulo $8.$

Edit: Even if you know that $kin1,2,3,4,5,6,7$ you can't necessarily do any better. Consider the number $10101010$ to see why.

share|cite|improve this answer

edited Jul 15 at 18:01

answered Jul 15 at 17:46

Cameron Buie

83.5k771153

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh you're right, I'll rephrase my question. I should've said that $k le 8$. Thanks.
  – Kratos
  Jul 15 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kratos: Unfortunately, that still doesn't allow you to do the job in general. See my edited answer.
  – Cameron Buie
  Jul 15 at 18:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm so dumb. Is there a way to calculate if k is odd or even?
  – Kratos
  Jul 15 at 18:10
  

  
  
  
  

add a comment | 

up vote
1
down vote

If you happen to start from 0101010101, you are out of luck because you cannot possible distinguish between $k=2$ and $k=0$ (and many more).
But if we assume that the original pattern has not periodic at all, the following works without trial shifting:

Note that circular shifting $a$ is equivalent to computing $2abmod 255$ (we exclude the case of $a=255$ here). Hence we are given $2^kabmod 255$ and want to find $k$. By assumption, $a$ is not a multiple of $17$ (or else the two half-bytes would be equal). Hence we can compute $a^-1bmod 17$ and hence $2^kbmod 17$. From this compute $k$ via lookup:
$$beginmatrix
2^k&1&2&4&8&9&13&15&16\
k&0&1&2&3&7&6&5&4endmatrix $$

share|cite|improve this answer

answered Jul 15 at 18:07

Hagen von Eitzen

265k20258477

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

You can't calculate $k$ at all, even by circularly shifting. Suppose you shifted it $j$ times (for some $jin0,1,2,3,4,5,6,7$ to get the original number. Then for any nonnegative integer $n,$ shifting the original number $8n-j$ times looks exactly the same as it would for any other such $n.$

The best you can do is calculate $k$ modulo $8.$

Edit: Even if you know that $kin1,2,3,4,5,6,7$ you can't necessarily do any better. Consider the number $10101010$ to see why.

share|cite|improve this answer

edited Jul 15 at 18:01

answered Jul 15 at 17:46

Cameron Buie

83.5k771153

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh you're right, I'll rephrase my question. I should've said that $k le 8$. Thanks.
  – Kratos
  Jul 15 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kratos: Unfortunately, that still doesn't allow you to do the job in general. See my edited answer.
  – Cameron Buie
  Jul 15 at 18:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm so dumb. Is there a way to calculate if k is odd or even?
  – Kratos
  Jul 15 at 18:10
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

You can't calculate $k$ at all, even by circularly shifting. Suppose you shifted it $j$ times (for some $jin0,1,2,3,4,5,6,7$ to get the original number. Then for any nonnegative integer $n,$ shifting the original number $8n-j$ times looks exactly the same as it would for any other such $n.$

The best you can do is calculate $k$ modulo $8.$

Edit: Even if you know that $kin1,2,3,4,5,6,7$ you can't necessarily do any better. Consider the number $10101010$ to see why.

share|cite|improve this answer

edited Jul 15 at 18:01

answered Jul 15 at 17:46

Cameron Buie

83.5k771153

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh you're right, I'll rephrase my question. I should've said that $k le 8$. Thanks.
  – Kratos
  Jul 15 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kratos: Unfortunately, that still doesn't allow you to do the job in general. See my edited answer.
  – Cameron Buie
  Jul 15 at 18:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm so dumb. Is there a way to calculate if k is odd or even?
  – Kratos
  Jul 15 at 18:10
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

You can't calculate $k$ at all, even by circularly shifting. Suppose you shifted it $j$ times (for some $jin0,1,2,3,4,5,6,7$ to get the original number. Then for any nonnegative integer $n,$ shifting the original number $8n-j$ times looks exactly the same as it would for any other such $n.$

The best you can do is calculate $k$ modulo $8.$

Edit: Even if you know that $kin1,2,3,4,5,6,7$ you can't necessarily do any better. Consider the number $10101010$ to see why.

share|cite|improve this answer

edited Jul 15 at 18:01

answered Jul 15 at 17:46

Cameron Buie

83.5k771153

You can't calculate $k$ at all, even by circularly shifting. Suppose you shifted it $j$ times (for some $jin0,1,2,3,4,5,6,7$ to get the original number. Then for any nonnegative integer $n,$ shifting the original number $8n-j$ times looks exactly the same as it would for any other such $n.$

The best you can do is calculate $k$ modulo $8.$

Edit: Even if you know that $kin1,2,3,4,5,6,7$ you can't necessarily do any better. Consider the number $10101010$ to see why.

share|cite|improve this answer

edited Jul 15 at 18:01

answered Jul 15 at 17:46

Cameron Buie

83.5k771153

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 18:01

edited Jul 15 at 18:01

edited Jul 15 at 18:01

answered Jul 15 at 17:46

Cameron Buie

83.5k771153

answered Jul 15 at 17:46

Cameron Buie

83.5k771153

answered Jul 15 at 17:46

Cameron Buie

83.5k771153

83.5k771153

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh you're right, I'll rephrase my question. I should've said that $k le 8$. Thanks.
  – Kratos
  Jul 15 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kratos: Unfortunately, that still doesn't allow you to do the job in general. See my edited answer.
  – Cameron Buie
  Jul 15 at 18:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm so dumb. Is there a way to calculate if k is odd or even?
  – Kratos
  Jul 15 at 18:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh you're right, I'll rephrase my question. I should've said that $k le 8$. Thanks.
  – Kratos
  Jul 15 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kratos: Unfortunately, that still doesn't allow you to do the job in general. See my edited answer.
  – Cameron Buie
  Jul 15 at 18:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm so dumb. Is there a way to calculate if k is odd or even?
  – Kratos
  Jul 15 at 18:10
  

  
  
  
  

Oh you're right, I'll rephrase my question. I should've said that $k le 8$. Thanks.
– Kratos
Jul 15 at 17:59

Oh you're right, I'll rephrase my question. I should've said that $k le 8$. Thanks.
– Kratos
Jul 15 at 17:59

@Kratos: Unfortunately, that still doesn't allow you to do the job in general. See my edited answer.
– Cameron Buie
Jul 15 at 18:02

@Kratos: Unfortunately, that still doesn't allow you to do the job in general. See my edited answer.
– Cameron Buie
Jul 15 at 18:02

I'm so dumb. Is there a way to calculate if k is odd or even?
– Kratos
Jul 15 at 18:10

I'm so dumb. Is there a way to calculate if k is odd or even?
– Kratos
Jul 15 at 18:10

add a comment | 

up vote
1
down vote

If you happen to start from 0101010101, you are out of luck because you cannot possible distinguish between $k=2$ and $k=0$ (and many more).
But if we assume that the original pattern has not periodic at all, the following works without trial shifting:

Note that circular shifting $a$ is equivalent to computing $2abmod 255$ (we exclude the case of $a=255$ here). Hence we are given $2^kabmod 255$ and want to find $k$. By assumption, $a$ is not a multiple of $17$ (or else the two half-bytes would be equal). Hence we can compute $a^-1bmod 17$ and hence $2^kbmod 17$. From this compute $k$ via lookup:
$$beginmatrix
2^k&1&2&4&8&9&13&15&16\
k&0&1&2&3&7&6&5&4endmatrix $$

share|cite|improve this answer

answered Jul 15 at 18:07

Hagen von Eitzen

265k20258477

add a comment | 

up vote
1
down vote

If you happen to start from 0101010101, you are out of luck because you cannot possible distinguish between $k=2$ and $k=0$ (and many more).
But if we assume that the original pattern has not periodic at all, the following works without trial shifting:

Note that circular shifting $a$ is equivalent to computing $2abmod 255$ (we exclude the case of $a=255$ here). Hence we are given $2^kabmod 255$ and want to find $k$. By assumption, $a$ is not a multiple of $17$ (or else the two half-bytes would be equal). Hence we can compute $a^-1bmod 17$ and hence $2^kbmod 17$. From this compute $k$ via lookup:
$$beginmatrix
2^k&1&2&4&8&9&13&15&16\
k&0&1&2&3&7&6&5&4endmatrix $$

share|cite|improve this answer

answered Jul 15 at 18:07

Hagen von Eitzen

265k20258477

add a comment | 

up vote
1
down vote

up vote
1
down vote

If you happen to start from 0101010101, you are out of luck because you cannot possible distinguish between $k=2$ and $k=0$ (and many more).
But if we assume that the original pattern has not periodic at all, the following works without trial shifting:

Note that circular shifting $a$ is equivalent to computing $2abmod 255$ (we exclude the case of $a=255$ here). Hence we are given $2^kabmod 255$ and want to find $k$. By assumption, $a$ is not a multiple of $17$ (or else the two half-bytes would be equal). Hence we can compute $a^-1bmod 17$ and hence $2^kbmod 17$. From this compute $k$ via lookup:
$$beginmatrix
2^k&1&2&4&8&9&13&15&16\
k&0&1&2&3&7&6&5&4endmatrix $$

share|cite|improve this answer

answered Jul 15 at 18:07

Hagen von Eitzen

265k20258477

If you happen to start from 0101010101, you are out of luck because you cannot possible distinguish between $k=2$ and $k=0$ (and many more).
But if we assume that the original pattern has not periodic at all, the following works without trial shifting:

Note that circular shifting $a$ is equivalent to computing $2abmod 255$ (we exclude the case of $a=255$ here). Hence we are given $2^kabmod 255$ and want to find $k$. By assumption, $a$ is not a multiple of $17$ (or else the two half-bytes would be equal). Hence we can compute $a^-1bmod 17$ and hence $2^kbmod 17$. From this compute $k$ via lookup:
$$beginmatrix
2^k&1&2&4&8&9&13&15&16\
k&0&1&2&3&7&6&5&4endmatrix $$

share|cite|improve this answer

answered Jul 15 at 18:07

Hagen von Eitzen

265k20258477

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 18:07

Hagen von Eitzen

265k20258477

answered Jul 15 at 18:07

Hagen von Eitzen

265k20258477

answered Jul 15 at 18:07

Hagen von Eitzen

265k20258477

265k20258477

add a comment | 

add a comment |