Spotify Codes are little barcodes that allow you to share songs, artists, users, playlists, etc.
They encode information in the different heights of the "bars". There are 8 discrete heights that the 23 bars can be, which means 8^23 different possible barcodes.
Spotify generates barcodes based on their URI schema. This URI spotify:playlist:37i9dQZF1DXcBWIGoYBM5M gets mapped to this barcode:
The URI has a lot more information (62^22) in it than the code. How would you map the URI to the barcode? It seems like you can't simply encode the URI directly. For more background, see my "answer" to this question: https://stackoverflow.com/a/62120952/10703868
The patent explains the general process, this is what I have found.
This is a more recent patent
When using the Spotify code generator the website makes a request to https://scannables.scdn.co/uri/plain/[format]/[background-color-in-hex]/[code-color-in-text]/[size]/[spotify-URI].
Using Burp Suite, when scanning a code through Spotify the app sends a request to Spotify's API: https://spclient.wg.spotify.com/scannable-id/id/[CODE]?format=json where [CODE] is the media reference that you were looking for. This request can be made through python but only with the [TOKEN] that was generated through the app as this is the only way to get the correct scope. The app token expires in about half an hour.
import requests
head={
"X-Client-Id": "58bd3c95768941ea9eb4350aaa033eb3",
"Accept-Encoding": "gzip, deflate",
"Connection": "close",
"App-Platform": "iOS",
"Accept": "*/*",
"User-Agent": "Spotify/8.5.68 iOS/13.4 (iPhone9,3)",
"Accept-Language": "en",
"Authorization": "Bearer [TOKEN]",
"Spotify-App-Version": "8.5.68"}
response = requests.get('https://spclient.wg.spotify.com:443/scannable-id/id/26560102031?format=json', headers=head)
print(response)
print(response.json())
Which returns:
<Response [200]>
{'target': 'spotify:playlist:37i9dQZF1DXcBWIGoYBM5M'}
So 26560102031 is the media reference for your playlist.
The patent states that the code is first detected and then possibly converted into 63 bits using a Gray table. For example 361354354471425226605 is encoded into 010 101 001 010 111 110 010 111 110 110 100 001 110 011 111 011 011 101 101 000 111.
However the code sent to the API is 6875667268, I'm unsure how the media reference is generated but this is the number used in the lookup table.
The reference contains the integers 0-9 compared to the gray table of 0-7 implying that an algorithm using normal binary has been used. The patent talks about using a convolutional code and then the Viterbi algorithm for error correction, so this may be the output from that. Something that is impossible to recreate whithout the states I believe. However I'd be interested if you can interpret the patent any better.
This media reference is 10 digits however others have 11 or 12.
Here are two more examples of the raw distances, the gray table binary and then the media reference:
1.
022673352171662032460
000 011 011 101 100 010 010 111 011 001 100 001 101 101 011 000 010 011 110 101 000
67775490487
2.
574146602473467556050
111 100 110 001 110 101 101 000 011 110 100 010 110 101 100 111 111 101 000 111 000
57639171874
edit:
Some extra info:
There are some posts online describing how you can encode any text such as spotify:playlist:HelloWorld into a code however this no longer works.
I also discovered through the proxy that you can use the domain to fetch the album art of a track above the code. This suggests a closer integration of Spotify's API and this scannables url than previously thought. As it not only stores the URIs and their codes but can also validate URIs and return updated album art.
https://scannables.scdn.co/uri/800/spotify%3Atrack%3A0J8oh5MAMyUPRIgflnjwmB
Your suspicion was correct - they're using a lookup table. For all of the fun technical details, the relevant patent is available here: https://data.epo.org/publication-server/rest/v1.0/publication-dates/20190220/patents/EP3444755NWA1/document.pdf
Very interesting discussion. Always been attracted to barcodes so I had to take a look. I did some analysis of the barcodes alone (didn't access the API for the media refs) and think I have the basic encoding process figured out. However, based on the two examples above, I'm not convinced I have the mapping from media ref to 37-bit vector correct (i.e. it works in case 2 but not case 1). At any rate, if you have a few more pairs, that last part should be simple to work out. Let me know.
For those who want to figure this out, don't read the spoilers below!
It turns out that the basic process outlined in the patent is correct, but lacking in details. I'll summarize below using the example above. I actually analyzed this in reverse which is why I think the code description is basically correct except for step (1), i.e. I generated 45 barcodes and all of them matched had this code.
1. Map the media reference as integer to 37 bit vector.
Something like write number in base 2, with lowest significant bit
on the left and zero-padding on right if necessary.
57639171874 -> 0100010011101111111100011101011010110
2. Calculate CRC-8-CCITT, i.e. generator x^8 + x^2 + x + 1
The following steps are needed to calculate the 8 CRC bits:
Pad with 3 bits on the right:
01000100 11101111 11110001 11010110 10110000
Reverse bytes:
00100010 11110111 10001111 01101011 00001101
Calculate CRC as normal (highest order degree on the left):
-> 11001100
Reverse CRC:
-> 00110011
Invert check:
-> 11001100
Finally append to step 1 result:
01000100 11101111 11110001 11010110 10110110 01100
3. Convolutionally encode the 45 bits using the common generator
polynomials (1011011, 1111001) in binary with puncture pattern
110110 (or 101, 110 on each stream). The result of step 2 is
encoded using tail-biting, meaning we begin the shift register
in the state of the last 6 bits of the 45 long input vector.
Prepend stream with last 6 bits of data:
001100 01000100 11101111 11110001 11010110 10110110 01100
Encode using first generator:
(a) 100011100111110100110011110100000010001001011
Encode using 2nd generator:
(b) 110011100010110110110100101101011100110011011
Interleave bits (abab...):
11010000111111000010111011110011010011110001...
1010111001110001000101011000010110000111001111
Puncture every third bit:
111000111100101111101110111001011100110000100100011100110011
4. Permute data by choosing indices 0, 7, 14, 21, 28, 35, 42, 49,
56, 3, 10..., i.e. incrementing 7 modulo 60. (Note: unpermute by
incrementing 43 mod 60).
The encoded sequence after permuting is
111100110001110101101000011110010110101100111111101000111000
5. The final step is to map back to bar lengths 0 to 7 using the
gray map (000,001,011,010,110,111,101,100). This gives the 20 bar
encoding. As noted before, add three bars: short one on each end
and a long one in the middle.
UPDATE: I've added a barcode (levels) decoder (assuming no errors) and an alternate encoder that follows the description above rather than the equivalent linear algebra method. Hopefully that is a bit more clear.
UPDATE 2: Got rid of most of the hard-coded arrays to illustrate how they are generated.
The linear algebra method defines the linear transformation (spotify_generator) and mask to map the 37 bit input into the 60 bit convolutionally encoded data. The mask is result of the 8-bit inverted CRC being convolutionally encoded. The spotify_generator is a 37x60 matrix that implements the product of generators for the CRC (a 37x45 matrix) and convolutional codes (a 45x60 matrix). You can create the generator matrix from an encoding function by applying the function to each row of an appropriate size generator matrix. For example, a CRC function that add 8 bits to each 37 bit data vector applied to each row of a 37x37 identity matrix.
import numpy as np
import crccheck
# Utils for conversion between int, array of binary
# and array of bytes (as ints)
def int_to_bin(num, length, endian):
if endian == 'l':
return [num >> i & 1 for i in range(0, length)]
elif endian == 'b':
return [num >> i & 1 for i in range(length-1, -1, -1)]
def bin_to_int(bin,length):
return int("".join([str(bin[i]) for i in range(length-1,-1,-1)]),2)
def bin_to_bytes(bin, length):
b = bin[0:length] + [0] * (-length % 8)
return [(b[i]<<7) + (b[i+1]<<6) + (b[i+2]<<5) + (b[i+3]<<4) +
(b[i+4]<<3) + (b[i+5]<<2) + (b[i+6]<<1) + b[i+7] for i in range(0,len(b),8)]
# Return the circular right shift of an array by 'n' positions
def shift_right(arr, n):
return arr[-n % len(arr):len(arr):] + arr[0:-n % len(arr)]
gray_code = [0,1,3,2,7,6,4,5]
gray_code_inv = [[0,0,0],[0,0,1],[0,1,1],[0,1,0],
[1,1,0],[1,1,1],[1,0,1],[1,0,0]]
# CRC using Rocksoft model:
# NOTE: this is not quite any of their predefined CRC's
# 8: number of check bits (degree of poly)
# 0x7: representation of poly without high term (x^8+x^2+x+1)
# 0x0: initial fill of register
# True: byte reverse data
# True: byte reverse check
# 0xff: Mask check (i.e. invert)
spotify_crc = crccheck.crc.Crc(8, 0x7, 0x0, True, True, 0xff)
def calc_spotify_crc(bin37):
bytes = bin_to_bytes(bin37, 37)
return int_to_bin(spotify_crc.calc(bytes), 8, 'b')
def check_spotify_crc(bin45):
data = bin_to_bytes(bin45,37)
return spotify_crc.calc(data) == bin_to_bytes(bin45[37:], 8)[0]
# Simple convolutional encoder
def encode_cc(dat):
gen1 = [1,0,1,1,0,1,1]
gen2 = [1,1,1,1,0,0,1]
punct = [1,1,0]
dat_pad = dat[-6:] + dat # 6 bits are needed to initialize
# register for tail-biting
stream1 = np.convolve(dat_pad, gen1, mode='valid') % 2
stream2 = np.convolve(dat_pad, gen2, mode='valid') % 2
enc = [val for pair in zip(stream1, stream2) for val in pair]
return [enc[i] for i in range(len(enc)) if punct[i % 3]]
# To create a generator matrix for a code, we encode each row
# of the identity matrix. Note that the CRC is not quite linear
# because of the check mask so we apply the lamda function to
# invert it. Given a 37 bit media reference we can encode by
# ref * spotify_generator + spotify_mask (mod 2)
_i37 = np.identity(37, dtype=bool)
crc_generator = [_i37[r].tolist() +
list(map(lambda x : 1-x, calc_spotify_crc(_i37[r].tolist())))
for r in range(37)]
spotify_generator = 1*np.array([encode_cc(crc_generator[r]) for r in range(37)], dtype=bool)
del _i37
spotify_mask = 1*np.array(encode_cc(37*[0] + 8*[1]), dtype=bool)
# The following matrix is used to "invert" the convolutional code.
# In particular, we choose a 45 vector basis for the columns of the
# generator matrix (by deleting those in positions equal to 2 mod 4)
# and then inverting the matrix. By selecting the corresponding 45
# elements of the convolutionally encoded vector and multiplying
# on the right by this matrix, we get back to the unencoded data,
# assuming there are no errors.
# Note: numpy does not invert binary matrices, i.e. GF(2), so we
# hard code the following 3 row vectors to generate the matrix.
conv_gen = [[0,1,0,1,1,1,1,0,1,1,0,0,0,1]+31*[0],
[1,0,1,0,1,0,1,0,0,0,1,1,1] + 32*[0],
[0,0,1,0,1,1,1,1,1,1,0,0,1] + 32*[0] ]
conv_generator_inv = 1*np.array([shift_right(conv_gen[(s-27) % 3],s) for s in range(27,72)], dtype=bool)
# Given an integer media reference, returns list of 20 barcode levels
def spotify_bar_code(ref):
bin37 = np.array([int_to_bin(ref, 37, 'l')], dtype=bool)
enc = (np.add(1*np.dot(bin37, spotify_generator), spotify_mask) % 2).flatten()
perm = [enc[7*i % 60] for i in range(60)]
return [gray_code[4*perm[i]+2*perm[i+1]+perm[i+2]] for i in range(0,len(perm),3)]
# Equivalent function but using CRC and CC encoders.
def spotify_bar_code2(ref):
bin37 = int_to_bin(ref, 37, 'l')
enc_crc = bin37 + calc_spotify_crc(bin37)
enc_cc = encode_cc(enc_crc)
perm = [enc_cc[7*i % 60] for i in range(60)]
return [gray_code[4*perm[i]+2*perm[i+1]+perm[i+2]] for i in range(0,len(perm),3)]
# Given 20 (clean) barcode levels, returns media reference
def spotify_bar_decode(levels):
level_bits = np.array([gray_code_inv[levels[i]] for i in range(20)], dtype=bool).flatten()
conv_bits = [level_bits[43*i % 60] for i in range(60)]
cols = [i for i in range(60) if i % 4 != 2] # columns to invert
conv_bits45 = np.array([conv_bits[c] for c in cols], dtype=bool)
bin45 = (1*np.dot(conv_bits45, conv_generator_inv) % 2).tolist()
if check_spotify_crc(bin45):
return bin_to_int(bin45, 37)
else:
print('Error in levels; Use real decoder!!!')
return -1
And example:
>>> levels = [5,7,4,1,4,6,6,0,2,4,3,4,6,7,5,5,6,0,5,0]
>>> spotify_bar_decode(levels)
57639171874
>>> spotify_barcode(57639171874)
[5, 7, 4, 1, 4, 6, 6, 0, 2, 4, 3, 4, 6, 7, 5, 5, 6, 0, 5, 0]
I have a testcase in which the behavior seems wrong. I see that in all generations the num_of_red_shoes is high, while I would expect a more even distribution. What is the cause of this behavior and how can it be fixed?
<'
struct closet {
kind:[SMALL,BIG];
num_of_shoes:uint;
num_of_red_shoes:uint;
num_of_black_shoes:uint;
num_of_yellow_shoes:uint;
keep soft num_of_red_shoes < 10;
keep soft num_of_black_shoes < 10;
keep soft num_of_yellow_shoes < 10;
keep num_of_yellow_shoes + num_of_black_shoes + num_of_red_shoes == num_of_shoes;
when BIG closet {
keep num_of_shoes in [50..100];
};
};
extend sys {
closets[100]:list of BIG closet;
};
'>
Generation results:
item type kind num_of_sh* num_of_re* num_of_bl* num_of_ye*
---------------------------------------------------------------------------
0. BIG closet BIG 78 73 1 4
1. BIG closet BIG 67 50 8 9
2. BIG closet BIG 73 68 0 5
3. BIG closet BIG 73 66 3 4
4. BIG closet BIG 51 50 0 1
5. BIG closet BIG 78 76 1 1
6. BIG closet BIG 55 43 7 5
7. BIG closet BIG 88 87 1 0
8. BIG closet BIG 99 84 6 9
9. BIG closet BIG 92 92 0 0
10. BIG closet BIG 63 55 3 5
11. BIG closet BIG 59 50 9 0
12. BIG closet BIG 51 44 2 5
13. BIG closet BIG 82 76 1 5
14. BIG closet BIG 81 74 2 5
15. BIG closet BIG 97 93 2 2
16. BIG closet BIG 54 41 8 5
17. BIG closet BIG 55 44 5 6
18. BIG closet BIG 70 55 9 6
19. BIG closet BIG 63 57 1 5
When there are contradicting soft constraints, Specman does not randomize the softs which are enforced, but rather gives priority to the constraints which were written last. Since the soft on red shoes was first in the test, it is the one which is always overridden.
If the softs are known to be mutually exclusive (which is not the case here) you could use a simple flag to randomly choose which soft should hold. e.g. the code would look like this:
flag:uint[0..2];
keep soft read_only(flag==0) => num_of_red_shoes < 10;
keep soft read_only(flag==1) => num_of_black_shoes < 10;
keep soft read_only(flag==2) => num_of_yellow_shoes < 10;
However, since here there is no knowledge in advance how many softs are expected to hold (and it's possible and two or all three will be satisfied) a more complex solution should be made. Here is a code which does this randomization:
struct closet {
kind:[SMALL,BIG];
num_of_shoes:uint;
num_of_red_shoes:uint;
num_of_black_shoes:uint;
num_of_yellow_shoes:uint;
//replaces the original soft constraints (if a flag is true the correlating
//right-side implication will be enforced
soft_flags[3]:list of bool;
keep for each in soft_flags {
soft it == TRUE;
};
//this list is used to shuffle the flags so their enforcement will be done
//with even distribution
soft_indices:list of uint;
keep soft_indices.is_a_permutation({0;1;2});
keep soft_flags[soft_indices[0]] => num_of_red_shoes < 10;
keep soft_flags[soft_indices[1]] => num_of_black_shoes < 10;
keep soft_flags[soft_indices[2]] => num_of_yellow_shoes < 10;
keep num_of_yellow_shoes + num_of_black_shoes + num_of_red_shoes == num_of_shoes;
};
I'm not with Cadence, so I can't give you a definite answer. I think the solver will try to break as few constraints as possible and it just chooses the first one if finds (in your case the one for red shoes). Try changing the order and see if this changes (if the black constraint is first, I'd think you'll always get more black shoes).
As a solution, you could just remove the soft constraints when you have a big closet:
when BIG closet {
keep num_of_red_shoes.reset_soft();
keep num_of_black_shoes.reset_soft();
keep num_of_yellow_shoes.reset_soft();
keep num_of_shoes in [50..100];
};
If you want to randomly choose which one of them to disable (sometimes more than 10 red shoes, sometimes more than 10 black shoes, etc.), then you'll need a helper field:
when BIG closet {
more_shoes : [ RED, BLACK, YELLOW ];
keep more_shoes == RED => num_of_red_shoes.reset_soft();
keep more_shoes == BLACK => num_of_black_shoes.reset_soft();
keep more_shoes == YELLOW => num_of_yellow_shoes.reset_soft();
keep num_of_shoes in [50..100];
};
It depends on what you mean by "a more even distribution".
There is no way to satisfy all of your hard and soft constraints for a BIG closet. Therefore Specman attempts to find a solution by ignoring some of your soft constraints. The IntelliGen constraint solver doesn't ignore all of the soft constraints, but attempts to find a solution while still using a subset. This is explained in the "Specman Generation User Guide" (sn_igenuser.pdf):
[S]oft constraints that are loaded later are considered to have a higher priority than soft constraints loaded previously."
In this case that means that Specman discards the soft constraint on red shoes and since it can find a solution still obeying the other soft constraints it does not discard them.
If you combine all of your soft constraints into one, then you will probably get the result you were hoping for:
keep soft ((num_of_red_shoes < 10) and (num_of_black_shoes < 10) and
(num_of_yellow_shoes < 10));
There are advantages to giving later constraints priority: This means that using AOP you can add new soft constraints and they will get the highest priority.
For more distributed values, I would suggest the following.
I'm sure you can follow the intended logic too.
var1, var2 : uint;
keep var1 in [0..30];
keep var2 in [0..30];
when BIG closet {
keep num_of_shoes in [50..100];
keep num_of_yellow_shoes == (num_of_shoes/3) - 15 + var1;
keep num_of_black_shoes == (num_of_shoes - num_of_yellow_shoes)/2 - 15 + var2;
keep num_of_red_shoes == num_of_shoes - (num_of_yellow_shoes - num_of_black_shoes);
keep gen (var1, var2) before (num_of_shoes);
keep gen (num_of_shoes) before (num_of_yellow_shoes, num_of_black_shoes, num_of_red_shoes);
keep gen (num_of_yellow_shoes) before (num_of_black_shoes, num_of_red_shoes);
keep gen (num_of_black_shoes) before (num_of_red_shoes);
};
I am new to Matlab and have been working my way through using Google. But now I have hit the wall it seems.
I have a text file which looks like following:
Information is for illustration reasons only
Aggregated Results
Date;$/Val1;Total $;Exp. Val1;Act. Val1
01-Oct-2008; -5.20; -1717; 330; 323
02-Oct-2008; -1.79; -595; 333; 324
03-Oct-2008; -2.29; -765; 334; 321
04-Oct-2008; -2.74; -917; 335; 317
Total Period; -0.80; -8612; 10748; 10276
Aggregated Results for location State PA
Date;$/Val1;Total $;Exp. Val1;Act. Val1
01-Oct-2008; -5.20; -1717; 330; 323
02-Oct-2008; -1.79; -595; 333; 324
03-Oct-2008; -2.29; -765; 334; 321
Total Period; -0.80; -8612; 10748; 10276
Results for account A1
Date;$/Val1;Total $;Exp. Val1;Act. Val1
01-Oct-2008; -7.59; -372; 49; 51
Total Period; -0.84; -1262; 1502; 1431
Results for account A2
Date;$/MWh;Total $;Exp. MWh;Act. MWh
01-Oct-2008; -8.00; -392; 49; 51
02-Oct-2008; 0.96; 47; 49; 51
03-Oct-2008; -0.75; -37; 50; 48
04-Oct-2008; 1.28; 53; 41; 40
Total Period; -0.36; -534; 1502; 1431
I want to extract following information in a cell/matrix format so that I can use it later to selectively do operations like average of accounts A1 and A2 or average of PA and A1, etc.
PA -0.8
A1 -0.84
A2 -0.036
I'd go this way:
fid = fopen(filename,'r');
A = textscan(fid,'%s','delimiter','\r');
A = A{:};
str_i = 'Total Period';
ix = find(strncmp(A,str_i,length(str_i)));
res = arrayfun(#(i) str2num(A{ix(i)}(length(str_i)+2:end)),1:numel(ix),'UniformOutput',false);
res = cat(2,res{:});
This way you'll get all the numerical values after a string 'Total Period' in a matrix, so that you may pick the values you need.
Similarly you may operate with strings PA, A1 and A2.
Matlab is not that nice when it comes to dealing with messy data. You may want to preprocess it a bit first.
However, here is an easy general way to import mixed numeric and non-numeric data in Matlab for a limited number of normal sized files.
Step 1: Copy the contents of the file into excel and save it as xls or xlsx
Step 2: Use xlsread
[NUM,TXT,RAW]=xlsread('test.xlsx')
From there the parsing should be maneagable.
Hopefully they will add non-numeric support to csvread or dlmread in the future.
Does anyone know how to calculate a Mod b in Casio fx-991ES Calculator. Thanks
This calculator does not have any modulo function. However there is quite simple way how to compute modulo using display mode ab/c (instead of traditional d/c).
How to switch display mode to ab/c:
Go to settings (Shift + Mode).
Press arrow down (to view more settings).
Select ab/c (number 1).
Now do your calculation (in comp mode), like 50 / 3 and you will see 16 2/3, thus, mod is 2. Or try 54 / 7 which is 7 5/7 (mod is 5).
If you don't see any fraction then the mod is 0 like 50 / 5 = 10 (mod is 0).
The remainder fraction is shown in reduced form, so 60 / 8 will result in 7 1/2. Remainder is 1/2 which is 4/8 so mod is 4.
EDIT:
As #lawal correctly pointed out, this method is a little bit tricky for negative numbers because the sign of the result would be negative.
For example -121 / 26 = -4 17/26, thus, mod is -17 which is +9 in mod 26. Alternatively you can add the modulo base to the computation for negative numbers: -121 / 26 + 26 = 21 9/26 (mod is 9).
EDIT2: As #simpatico pointed out, this method will not work for numbers that are out of calculator's precision. If you want to compute say 200^5 mod 391 then some tricks from algebra are needed. For example, using rule
(A * B) mod C = ((A mod C) * B) mod C we can write:
200^5 mod 391 = (200^3 * 200^2) mod 391 = ((200^3 mod 391) * 200^2) mod 391 = 98
As far as I know, that calculator does not offer mod functions.
You can however computer it by hand in a fairly straightforward manner.
Ex.
(1)50 mod 3
(2)50/3 = 16.66666667
(3)16.66666667 - 16 = 0.66666667
(4)0.66666667 * 3 = 2
Therefore 50 mod 3 = 2
Things to Note:
On line 3, we got the "minus 16" by looking at the result from line (2) and ignoring everything after the decimal. The 3 in line (4) is the same 3 from line (1).
Hope that Helped.
Edit
As a result of some trials you may get x.99991 which you will then round up to the number x+1.
You need 10 ÷R 3 = 1
This will display both the reminder and the quoitent
÷R
There is a switch a^b/c
If you want to calculate
491 mod 12
then enter 491 press a^b/c then enter 12. Then you will get 40, 11, 12. Here the middle one will be the answer that is 11.
Similarly if you want to calculate 41 mod 12 then find 41 a^b/c 12. You will get 3, 5, 12 and the answer is 5 (the middle one). The mod is always the middle value.
You can calculate A mod B (for positive numbers) using this:
Pol( -Rec( 1/2πr , 2πr × A/B ) , Y ) ( πr - Y ) B
Then press [CALC], and enter your values for A and B, and any value for Y.
/ indicates using the fraction key, and r means radians ( [SHIFT] [Ans] [2] )
type normal division first and then type shift + S->d
Here's how I usually do it. For example, to calculate 1717 mod 2:
Take 1717 / 2. The answer is 858.5
Now take 858 and multiply it by the mod (2) to get 1716
Finally, subtract the original number (1717) minus the number you got from the previous step (1716) -- 1717-1716=1.
So 1717 mod 2 is 1.
To sum this up all you have to do is multiply the numbers before the decimal point with the mod then subtract it from the original number.
Note: Math error means a mod m = 0
It all falls back to the definition of modulus: It is the remainder, for example, 7 mod 3 = 1.
This because 7 = 3(2) + 1, in which 1 is the remainder.
To do this process on a simple calculator do the following:
Take the dividend (7) and divide by the divisor (3), note the answer and discard all the decimals -> example 7/3 = 2.3333333, only worry about the 2. Now multiply this number by the divisor (3) and subtract the resulting number from the original dividend.
so 2*3 = 6, and 7 - 6 = 1, thus 1 is 7mod3
Calculate x/y (your actual numbers here), and press a b/c key, which is 3rd one below Shift key.
Simply just divide the numbers, it gives yuh the decimal format and even the numerical format. using S<->D
For example: 11/3 gives you 3.666667 and 3 2/3 (Swap using S<->D).
Here the '2' from 2/3 is your mod value.
Similarly 18/6 gives you 14.833333 and 14 5/6 (Swap using S<->D).
Here the '5' from 5/6 is your mod value.
I have a program output with one tandem repeat in different variants. Is it possible to search (in a string) for the motif and to tell the program to find all variants with maximum "3" mismatches/insertions/deletions?
I will take a crack at this with the very limited information supplied.
First, a short friendly editorial:
<editorial>
Please learn how to ask a good question and how to be precise.
At a minimum, please:
Refrain from domain specific jargon such as "motif" and "tandem repeat" and "base pairs" without providing links or precise definitions;
Say what the goal is and what you have done so far;
Important: Provide clear examples of input and desired output.
It is not helpful to potential helpers on SO have to have to play 20 questions in comments to try and understand your question! I spent more time trying to figure out what you were asking than answering it.
</editorial>
The following program generates a string of 2 character pairs 5,428 pairs long in an array of 1,000 elements long. I realize it is more likely that you will be reading these from a file, but this is just an example. Obviously you would replace the random strings with your actual data from whatever source.
I do not know if 'AT','CG','TC','CA','TG','GC','GG' that I used are legitimate base pair combinations or not. (I slept through biology...) Just edit the map block pairs to legitimate pairs and change the 7 to the number of pairs if you want to generate legitimate random strings for testing.
If the substring at the offset point is 3 differences or less, the array element (a scalar value) is stored in an anonymous array in the value part of a hash. The key part of the hash is the substring that is a near match. Rather than array elements, the values could be file names, Perl data references or other relevant references you want to associate with your motif.
While I have just looked at character by character differences between the strings, you can put any specific logic that you need to look at by replacing the line foreach my $j (0..$#a1) { $diffs++ unless ($a1[$j] eq $a2[$j]); } with the comparison logic that works for your problem. I do not know how mismatches/insertions/deletions are represented in your string, so I leave that as an exercise to the reader. Perhaps Algorithm::Diff or String::Diff from CPAN?
It is easy to modify this program to have keyboard input for $target and $offset or have the string searched beginning to end rather than several strings at a fixed offset. Once again: it was not really clear what your goal is...
use strict; use warnings;
my #bps;
push(#bps,join('',map { ('AT','CG','TC','CA','TG','GC','GG')[rand 7] }
0..5428)) for(1..1_000);
my $len=length($bps[0]);
my $s_count= scalar #bps;
print "$s_count random strings generated $len characters long\n" ;
my $target="CGTCGCACAG";
my $offset=832;
my $nlen=length $target;
my %HoA;
my $diffs=0;
my #a2=split(//, $target);
substr($bps[-1], $offset, $nlen)=$target; #guarantee 1 match
substr($bps[-2], $offset, $nlen)="CATGGCACGG"; #anja example
foreach my $i (0..$#bps) {
my $cand=substr($bps[$i], $offset, $nlen);
my #a1=split(//, $cand);
$diffs=0;
foreach my $j (0..$#a1) { $diffs++ unless ($a1[$j] eq $a2[$j]); }
next if $diffs > 3;
push (#{$HoA{$cand}}, $i);
}
foreach my $hit (keys %HoA) {
my #a1=split(//, $hit);
$diffs=0;
my $ds="";
foreach my $j (0..$#a1) {
if($a1[$j] eq $a2[$j]) {
$ds.=" ";
} else {
$diffs++;
$ds.=$a1[$j];
}
}
print "Target: $target\n",
"Candidate: $hit\n",
"Differences: $ds $diffs differences\n",
"Array element: ";
foreach (#{$HoA{$hit}}) {
print "$_ " ;
}
print "\n\n";
}
Output:
1000 random strings generated 10858 characters long
Target: CGTCGCACAG
Candidate: CGTCGCACAG
Differences: 0 differences
Array element: 999
Target: CGTCGCACAG
Candidate: CGTCGCCGCG
Differences: CGC 3 differences
Array element: 696
Target: CGTCGCACAG
Candidate: CGTCGCCGAT
Differences: CG T 3 differences
Array element: 851
Target: CGTCGCACAG
Candidate: CGTCGCATGG
Differences: TG 2 differences
Array element: 986
Target: CGTCGCACAG
Candidate: CATGGCACGG
Differences: A G G 3 differences
Array element: 998
..several cut out..
Target: CGTCGCACAG
Candidate: CGTCGCTCCA
Differences: T CA 3 differences
Array element: 568 926
I believe that there are routines for this sort of thing in BioPerl.
In any case, you might get better answers if you asked this over at BioStar, the bioinformatics stack exchange.
When I was in my first couple years of learning perl, I wrote what I now consider to be a very inefficient (but functional) tandem repeat finder (which used to be available on my old job's company website) called tandyman. I wrote a fuzzy version of it a couple years later called cottonTandy. If I were to re-write it today, I would use hashes for a global search (given the allowed mistakes) and utilize pattern matching for a local search.
Here's an example of how you use it:
#!/usr/bin/perl
use Tandyman;
$sequence = "ATGCATCGTAGCGTTCAGTCGGCATCTATCTGACGTACTCTTACTGCATGAGTCTAGCTGTACTACGTACGAGCTGAGCAGCGTACgTG";
my $tandy = Tandyman->new(\$sequence,'n'); #Can't believe I coded it to take a scalar reference! Prob. fresh out of a cpp class when I wrote it.
$tandy->SetParams(4,2,3,3,4);
#The parameters are, in order:
# repeat unit size
# min number of repeat units to require a hit
# allowed mistakes per unit (an upper bound for "mistake concentration")
# allowed mistakes per window (a lower bound for "mistake concentration")
# number of units in a "window"
while(#repeat_info = $tandy->FindRepeat())
{print(join("\t",#repeat_info),"\n")}
The output of this test looks like this (and takes a horrendous 11 seconds to run):
25 32 TCTA 2 0.87 TCTA TCTG
58 72 CGTA 4 0.81 CTGTA CTA CGTA CGA
82 89 CGTA 2 0.87 CGTA CGTG
45 51 TGCA 2 0.87 TGCA TGA
65 72 ACGA 2 0.87 ACGT ACGA
23 29 CTAT 2 0.87 CAT CTAT
36 45 TACT 3 0.83 TACT CT TACT
24 31 ATCT 2 1 ATCT ATCT
51 59 AGCT 2 0.87 AGTCT AGCT
33 39 ACGT 2 0.87 ACGT ACT
62 72 ACGT 3 0.83 ACT ACGT ACGA
80 88 ACGT 2 0.87 AGCGT ACGT
81 88 GCGT 2 0.87 GCGT ACGT
63 70 CTAC 2 0.87 CTAC GTAC
32 38 GTAC 2 0.87 GAC GTAC
60 74 GTAC 4 0.81 GTAC TAC GTAC GAGC
23 30 CATC 2 0.87 CATC TATC
71 82 GAGC 3 0.83 GAGC TGAGC AGC
1 7 ATGC 2 0.87 ATGC ATC
54 60 CTAG 2 0.87 CTAG CTG
15 22 TCAG 2 0.87 TCAG TCGG
70 81 CGAG 3 0.83 CGAG CTGAG CAG
44 50 CATG 2 0.87 CTG CATG
25 32 TCTG 2 0.87 TCTA TCTG
82 89 CGTG 2 0.87 CGTA CGTG
55 73 TACG 5 0.75 TAGCTG TAC TACG TACG AG
69 83 AGCG 4 0.81 ACG AGCTG AGC AGCG
15 22 TCGG 2 0.87 TCAG TCGG
As you can see, it allows indels and SNPs. The columns are, in order:
Start position
Stop position
Consensus sequence
The number of units found
A quality metric out of 1
The repeat units separated by spaces
Note, that it's easy to supply parameters (as you can see from the output above) that will output junk/insignificant "repeats", but if you know how to supply good params, it can find what you set it upon finding.
Unfortunately, the package is not publicly available. I never bothered to make it available since it's so slow and not amenable to even prokaryotic-sized genome searches (though it would be workable for individual genes). In my novice coding days, I had started to add a feature to take a "state" as input so that I could run it on sections of a sequence in parallel and I never finished that once I learned hashes would make it so much faster. By that point, I had moved on to other projects. But if it would suit your needs, message me, I can email you a copy.
It's just shy of 1000 lines of code, but it has lots of bells & whistles, such as the allowance of IUPAC ambiguity codes (BDHVRYKMSWN). It works for both amino acids and nucleic acids. It filters out internal repeats (e.g. does not report TTTT or ATAT as 4nt consensuses).