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Distinct sum of 3 variable which ranges between 1-255

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Ahsen Noor Subhopoto
Ahsen Noor Subhopoto 2019 年 11 月 9 日
Edited: John D'Errico 2019 年 11 月 15 日
Hi, I am trying to retrieve array of sum of 3 numbers (x,y,z) each has a range of 1-255, in such a way that each sum is different from others. I have coded but I couldn't get 100% distinct sum, below result gives Average = 255, Max = 255 , MIN = (254 but I want 255 every time if possible). So each variable's value can be decoded back form the distinct sum. For example sum = 3 is only possible when we have (1+1+1) but for Sum = 4 we have 3 different possible and then it is not decodable; (1+1+2),(1+2+1),(2+1+1), sum = 4. I would be happy if anyone of you can debug this problem. Thank you.
%Updated.
%Input and Expected Output%
% Input Output
% [1,1,2] sum=>4
% [1,2,1] sum=>[different than above]
% [2,1,1] sum=>[distinct from others]
% [3,1,1] sum=>5
% ... ...
%Code%
clear;clc;
for n = 1:1000 %using Iteration to have 1000 different results that are stored in Difference(n,:) which gives Uniqueness values of each result.
Array = zeros(255,5);
Array(:,1:3) = randi(255,255,3);
for k = 1:length(Array);
Array(k,4) = sum(Array(k,1:3));
end
for l = 1:length(Array);
firstVal = Array(l,1)*0.7;
secondVal = Array(l,2)*0.9+firstVal;
thirdVal = Array(l,3)*1.4*secondVal;
Array(l,5) = sum([firstVal,secondVal,thirdVal]);
end
%253Avg
% first = Array(l,1)*0.7;
% second = Array(l,2)*0.9 + first;
% third = Array(l,3)*1.4 + second;
oldunique_Length = length(unique(Array(:,4)));
newunique_Length = length(unique(Array(:,5)));
Difference(n,:) = [oldunique_Length,newunique_Length];
end
%average no. of distinct sum = 255/255, Max value is 255 and MIN value is 254. It is still not 100% because I am getting 254 as a MIN. Everytime I should be able to get 255 (all rows must be different from others)
["OLD UniqueIds (AVG,MAX,MIN)",round(mean(Difference(:,1))),max(Difference(:,1)),min(Difference(:,1));"NEW UniqueIds (AVG,MAX,MIN)",round(mean(Difference(:,2))),max(Difference(:,2)),min(Difference(:,2))]

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採用された回答

John D'Errico
John D'Errico 2019 年 11 月 15 日
編集済み: John D'Errico 2019 年 11 月 15 日
As you recognize yourself, if you simply form the sum of three integers, you cannot then recover the original values from only that sum. Sufficient information is lost to recover the original components. Thus, given
2 + 1 + 1 == 1 + 2 + 1 == 1 + 1 + 2 == 4
if the sum was 4, then it is never possible to identify which of the three options caused that sum. Sorry, but that is simple fact. You cannot escape the laws of mathematics.
However, you CAN do something that allows recovery, if you use a proper encoding. Think of it as a base transformation, working in base 255. Given three integers, each in the range 1 - 255. I'll call them K, N, & M. Form an encoding as:
KNM = 255^2*(K-1) + 255*(N-1) + (M-1)
Thus for any values of K, N, M, we will get a distinct result, treating those numbers as effectively "digits" in base 255, then mapping the result to base 10. Try it out.
KNMencode = @(K,N,M) 255^2*(K-1) + 255*(N-1) + (M-1);
KNMencode(2,1,1)
ans =
65025
KNMencode(1,2,1)
ans =
255
KNMencode(1,1,2)
ans =
1
So we now get a distinct and unique result for any input combination in that range.
The reverse is true too. Now you can decode any possible integer value in the interval [0,16451834], back into the original set that created the number.
KNMencode(1,1,1)
ans =
0
KNMencode(255,255,255)
ans =
16451834
Again, this is just a case of a base conversion from base 10 back into base 255. Lets try it out.
KNMdecode = @(E) [mod(floor(E/255^2),255) + 1,mod(floor(E/255),255) + 1,mod(E,255) + 1];
KNMencode(1,2,3)
ans =
257
KNMdecode(257)
ans =
1 2 3
So any triple of integers (K,N,M) will result in a unique encoded value as needed, that can then be decoded, recovering the original values.

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More Answers (1)

Praveen Iyyappan Valsala
Praveen Iyyappan Valsala 2019 年 11 月 9 日
I didn't understand your way for making the sum unique. This dirty code gives you 255 (x,y,z) whose sums are unique.
sumArr=-1*ones(1,255); %Initialize with impossible sums
Array1=zeros(255,3);
for i=1:255
guess=randi([1 255],[1 3]);
while(sum(sumArr==sum(guess))~=0)
guess=randi(255,[1 3]); % guess unil sum is unique
end
sumArr(i)=sum(guess);
Array1(i,:)=guess;
end
if you just need (sum) array which is unique, use randperm instead of randi
unique_sumArr=randperm(255*3,255); %adjust the range

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Ahsen Noor Subhopoto
Ahsen Noor Subhopoto 2019 年 11 月 10 日
  1. Correction: With a single known I want to have 3 unknowns which I think is possible when we can manipulate x,y,z with some constants because 3 unknowns are basically bounded by the range of 1-255.
In the above code which I have updated you will see the Array(:,5) is distinct sum array for each x,y,z.
Praveen Iyyappan Valsala
Praveen Iyyappan Valsala 2019 年 11 月 11 日
Then why don't use something like.
Array(:,5)=Array(:,1)+Array(:,2)*1000+Array(:,3)*1000000;
I'm no expert in math. one way to find how many unique sums your expression can produce is by brute force. It is possible if your have enough memory(~16GB).
%permutation of all (x,y,z) with repetition
[x, y ,z]=ndgrid(1:255);
Array=[x(:)*0.7 y(:)*0.9+x(:)*0.7 z(:)*1.4.*(y(:)*0.9+x(:)*0.7 )];
sum_array=sum(Array,2).';
%sort for run length encoding
[sum_array,idx]=sort(sum_array);
Array=Array(idx,:);
% find unique elements by rle
% https://in.mathworks.com/matlabcentral/fileexchange/4955-rle-de-encoding
i = [ find(sum_array(1:end-1) ~= sum_array(2:end)) length(sum_array) ];
Nrep = diff([ 0 i ]);
elem = sum_array(i);
unique_sum=elem(Nrep==1);
xyz=Array(i,:);
unique_xyz=xyz(Nrep==1,:);
%check
assert(isequal(sum(unique_xyz,2).',unique_sum))
out of 16581375 permutation you have only 4278778 unique sums for your expression
Finally, don't use unique function. That is not what you need.
unique([1 1 1 2 3 4])
>> [1 2 3 4] % output of unique: not what you need
>> [ 2 3 4] % what you neeed
Ahsen Noor Subhopoto
Ahsen Noor Subhopoto 2019 年 11 月 15 日
Thank you for your time. I will look at it and other different techniques to make it 100% unique values.

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