#1




Find the Algorithm
I would like to see some of my friends take the algorithm out of these formulas
They all count on one algorithm 0101=70f5 de50=2c8c 0102=dde2 ffff5967 
#2




An infinite number of mathematical or algorithmic mappings are possible. Probably you need to be more specific on what operations are allowed and maximum number of ops as well as maybe more values. Otherwise easy enough:
(x&0xf==1)*0x70f5+(x&0xf==0)*0x2c8c+(x&0xf==2)*0xdde2+(x&0xf==0xf)*0x5967 Technically speaking that one liner is an algorithm upon which they all rely. I'm not sure how to properly mathematically phrase what I feel you ar asking, as doubtlessly not my silly example, but it's interesting in how hard the question might be to exactly specify. I would think something like the minimal kolmogorov complexity but even then maybe this formula is simpler with a mere 4 cases. I would like to see not only the intended answer but how are we to ask in a way which excludes all the endless wrong ones 
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mcr4ck (01142020) 
#3




I didn't understand anything from the formula
I think it's flawed On the other hand, the numbers that are written together are calculated by a formula That is, they are separate 0101=70f5 or de50=2c8c or 0102=dde2 or ffff=5967 Here's the formula for calculating this That is to say, if we give an input to these numbers, give me an output according to the numbers above. for example: in 0101 ==> out 70f5 or in de50 ==> out 2c8c or in 0102 ==> out dde2 
#4




My formula used bit wise and &, equality comparison ==, implicit cast of true/false to 1/0 and integer multiplication *, and integer addition +. It should work in C or Javascript or what have you. x is the input, the result is the output. Exactly as you specify it.
This works since the last digits are all unique. By bitwise and with 0xf, we get the last digit of the input and compare with each unique last digit. If equal it will result in 1 times the correct output value otherwise 0. Then summed all together maps them properly. Any one to one mapping can be done without the bitwise and, using only the other 4 operations. Of course it will only work for individual inputs then and will always be zero for not yet seen numbers. Currently it works for individual input last digits that match and is 0 for all other inputs. Last edited by chants; 01142020 at 04:44. 
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mcr4ck (01142020) 
#5




I didn't really understand what you meant
Can you give an example It means giving in and taking out 
#6




Just open a Python 3.7 console:
Code:
>>> x=0x0101 >>> hex((x&0xf==1)*0x70f5+(x&0xf==0)*0x2c8c+(x&0xf==2)*0xdde2+(x&0xf==0xf)*0x5967) '0x70f5' >>> x=0xde50 >>> hex((x&0xf==1)*0x70f5+(x&0xf==0)*0x2c8c+(x&0xf==2)*0xdde2+(x&0xf==0xf)*0x5967) '0x2c8c' >>> x=0x0102 >>> hex((x&0xf==1)*0x70f5+(x&0xf==0)*0x2c8c+(x&0xf==2)*0xdde2+(x&0xf==0xf)*0x5967) '0xdde2' >>> x=0xffff >>> hex((x&0xf==1)*0x70f5+(x&0xf==0)*0x2c8c+(x&0xf==2)*0xdde2+(x&0xf==0xf)*0x5967) '0x5967' Code:
f(x)=(x&0xf==1)*0x70f5+(x&0xf==0)*0x2c8c+(x&0xf==2)*0xdde2+(x&0xf==0xf)*0x5967 
#7




Too many codes I just cited, for example
You should not relate to each other Because I want to get the output Now if I change the input the output will change too I want a single formula that matches the input and output I gave That means the number of entries is too high, and I only said eight You have included all four outputs in the calculation in 0202===> out ? 
#8




What chants is saying, is that you can't derive the algorithm from 4 challengeresponse combinations.
Even if you had a lot more combinations of challengeresponse codes, it would probably be difficult to deduce the formula. The formula chants provided works with the four combinations you provided. It does not mean that it is the formula that originally gave them. This means that if you had a fifth pair of challengeresponse codes, the output derived from chant's formula would almost certainly mismatch the responce code of your blackbox. 
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chants (01202020) 
#9




hi
for more example: 3300 > 729C 3301 > BEF5 3302 > 4BE2 3303 > 111B 3304 > 2750 3305 > 7DA9 3306 > 21A6 3307 > AEDF 3308 > E594 3309 > 77ED 330A > 805A 330B > 4993 330C > A148 330D > D121 330E > B6DE 330F > 31D7 3310 > 2DCC 3311 > 5FA5 3312 > 6352 3313 > F40B 3314 > AD80 3315 > B999 3316 > C356 3317 > 610F 3318 > 2804 3319 > 5A9D 331A > 348A 331B > 9403 331C > 6DF8 331D > 9851 Can we make an algorithm for these? 
#10




hi
for more example: 3300 > 729C 3301 > BEF5 3302 > 4BE2 3303 > 111B 3304 > 2750 3305 > 7DA9 3306 > 21A6 3307 > AEDF 3308 > E594 3309 > 77ED 330A > 805A 330B > 4993 330C > A148 330D > D121 330E > B6DE 330F > 31D7 3310 > 2DCC 3311 > 5FA5 3312 > 6352 3313 > F40B 3314 > AD80 3315 > B999 3316 > C356 3317 > 610F 3318 > 2804 3319 > 5A9D 331A > 348A 331B > 9403 331C > 6DF8 331D > 9851 Can we make an algorithm for these? 
#11




yes, obligate/set anytine 331D as enter values and you will know anytime the answer
__________________
Decode and Conquer 
#12




Can you explain by example
I want to find the algorithm constant That is, give the input high numbers and deliver the output that is written in front of it. for example: in out 3300 ==> 729C or 3301==>BEF5 or 3302==>4BE2 
#13




Well the last bit of numbers never change so
Odd > odd, even > even, in fact even last 2 bits are constant, so modulo 4 constant and simple mapping of the other 2 bits. Also the last 4 bits are easy one to one mapping 0 to C 1 to 5 2 to 2 3 to B 4 to 0 5 to 9 6 to 6 7 to F 8 to 4 9 to D A to A B to 3 C to 8 D to 1 E to E F to 7 So remainder 2 stays same, remainder 0 adds C, remainder 1 adds 4, remainder 3 adds 8, all of these modulo 16, to keep 4 bits. Easy to write as a formula in a variety of ways from boolean equations to division and modulo or doing some bit twiddling. I don't see how the first 12 bits are derived though 
#14




Please give an example
I did not catch 
#15




Its the formula for the last hex digit. I don't know how the first 3 digits are computed.
Code:
((input%16)+((input%4)<=1)*4+(((input1)%4)>=2)*8)%16 
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