SQL DATA BITWISE

Operators

 1 & 1 equal 1
0 & 0 equal 0
1 & 0 equal 0
& only return 1 when both are 1

0 | 1 equal 1
0 | 0 equal 0
1 | 0 equal 1
| only return 1 when either is 1

0 ^ 0 equal 0
0 ^ 1 equal 1
1 ^ 1 equal 1
^ only return 1 if both is different

 ~ 0 equal 1
 ~ 1 equal  0
~ flip the coin 0 to 1 or 1 to 0

<< 1 shift left  will add 0 push number << but remain same number of digits.
>> 1 shift right will add 0 push >> but remain same number of digits


BITWISE  &

As bitwise is like converting a into binary then using operator add or xor or nor to get a different bit , can return as decimal or stay as bits. Or adding 1 zero behind a binary then data that is retreive is different so
A in sql after call from table.
I can add bitwise to get a different value
but the storing of A is still A no different in table data. Adding more column
looking for button F so
let x = 'a'.charCodeAt(0).toString(2) & 'f'.charCodeAt(0).toString(2);
x is 1099840  = char a & char f

But keyboard f will distinguish action button, solo f won't change to chopping to axing stone.
So i think the forest tree need to do converstion or walkman
doing converstion need basic  grab
char doing standing       A     &  F button  & resource a
                walking        97      &  F & a
                   jumping     61     & F &  a
                 squad down141    ??basically  not choping or axing so first 3 will do

It seems i ran into same number for a & f & a and a & f & b which both is 1099840 which cannot do for chopping and axing will be same picture

so this theory will be fail.

Column1 Hexaldecimal octal Decimal Binary Bota function Bitwise a & f & a Bitwise  97 & f & a Bitwise 61 & f & a Bitwise
a & f & b
Bitwise 97 & f & b Bitwise 61 & f & b
_ 5F 137 95 1011111 Xw==            
a 61 141 97 1100001 YQ==   1099840 32768 34816 1099840 32776 34826
b 62 142 98 1100010 Yg==            
c 63 143 99 1100011 Yw==            
d 64 144 100 1100100 ZA==            
e 65 145 101 1100101 ZQ==            
f 66 146 102 1100110 Zg==            
g 67 147 103 1100111 Zw==            
h 68 150 104 1101000 aA==            

BITWISE >>

so this theory will be fail.

so i try << shifting

let x = 'a'.charCodeAt(0).toString(2) << 'f'.charCodeAt(0).toString(2) ;
let x2 = x << 'a'.charCodeAt(0).toString(2);

And it ended up with suprise
a  << f  get 842547200
a << f << a get 1685094400

a << f << b  -520093696
a <<f << c -1040187392
so i just need to auto remove - sign  Math.abs(x)

And it work all unique value to put xxx.png and will show different picture when chopping and axing stone

But when put to B character it show the same number for few 97 << f << a and 98 <<f<<a  and 61 <<f<<a  62 <<f<<a

still fail

Column1 Hexaldecimal octal Decimal Binary Bota function Bitwise a << f <<a Bitwise  97 << f << a Bitwise 61 << f << a Bitwise
a << f << b
Bitwise 97 << f << b Bitwise 61 << f << b
_ 5F 137 95 1011111 Xw==            
a 61 141 97 1100001 YQ==   1685094400 657686528 686882816 520093696 1728053248 503316480
b 62 142 98 1100010 Yg== 1685389312 657686528 686882816 369098752 1728053248 503316480
c 63 143 99 1100011 Yw==            
d 64 144 100 1100100 ZA==            
/ 2F 57 47 101111 Lw==            
0 30 60 48 110000 MA==            

BITWISE >> ^

so this theory will be fail.

so i try << shifting

let x = 'a'.charCodeAt(0).toString(2) ^ 'f'.charCodeAt(0).toString(2) ;
let x2 = x << 'a'.charCodeAt(0).toString(2);

And it ended up with suprise
a  ^ f    a  get 862
a  ^ f << a  get 862

a ^f  <<b  441344
a ^f  <<c   882688


so i just need to auto remove - sign  Math.abs(x)

And it work all unique value to put xxx.png and will show different picture when chopping and axing stone

But when put to B character it show the same number for few 97 ^ f << a and 98 ^f<<a  and 61 ^f<<a  62 ^f<<a

still fail

Column1 Hexaldecimal octal Decimal Binary Bota function Bitwise a ^ f <<a Bitwise  97 ^ f << a Bitwise 61 ^ f << a Bitwise
a ^ f << b
Bitwise 97 ^ f << b Bitwise 61 ^ f << b
[ 5B 133 91 1011011 Ww==            
\ 5C 134 92 1011100 XA==            
] 5D 135 93 1011101 XQ==            
^ 5E 136 94 1011110 Xg==            
_ 5F 137 95 1011111 Xw==            
a 61 141 97 1100001 YQ==   862 2290094 2281120 441344 1172528128 1167933440
b 62 142 98 1100010 Yg== 840 2290094 2281120 430080 1172528128 1167933440
0 30 60 48 110000 MA==            

BITWISE >> ^

So everything fail so i just do the first if habitat.convert()
Tree will do this formula
tree name is 'a' convert to binaray with ^ f button as f is the keyboard

let x = 'a'.charCodeAt(0).toString(2) ^ 'f'.charCodeAt(0).toString(2) ;

it seems that decimal ^ f will get same number no result

 seems that hexa ^ f will get same number no result

Also octal ^ will get same number no result.

binaray ^ f  give unique

if habitat.convert()
rock will do another formula
 then my habitat need do switch more coding. Can only do straight if naming is a or b it bitwisse ^ with keyboard 'f' as compute is after press 'f'


Column1 Hexaldecimal octal Decimal Binary Bota function Bitwise key ^ f Bitwise  key & f Bitwise hex ^ f Bitwise
oct ^ f
 key ^ f + walkman naming
[ 5B 133 91 1011011 Ww== 2073613 18754 1140571 1139967  
\ need \\ sowon't use 5C 134 92 1011100 XA== 2073810 18700 1140571 1139967  
] 5D 135 93 1011101 XQ== 2073811 18700 1140571 1139967  
^ 5E 136 94 1011110 Xg== 2073832 18694 1140571 1139967  
_ 5F 137 95 1011111 Xw== 2073833 18694 1140571 1139967  
a 61 141 97 1100001 YQ==   431 1099840 1140560 1139967 43161
431141
43197
if b resouce
420a
42061
420141
42097
b 62 142 98 1100010 Yg== 420 1099850 1140560   420a
42061
420141
42097
c 63 143 99 1100011 Yw== 421 1099850 1140560    
d 64 144 100 1100100 ZA== 10 1100100 1140560    
j 6A 152 106 1101010 ag== 1436       6Aa
152a
106a

BITWISE >> ^

So everything fail so i just do the first if habitat.convert()
Tree will do this formula
tree name is 'a' convert to binaray with ^ f button as f is the keyboard

let x = 'a'.charCodeAt(0).toString(2) ^ 'f'.charCodeAt(0).toString(2) ;


So after getting unique 431 i pass to walkman.doconvert again

inside will be also a but bitwise again ?
seems 431 bitwise a again will hit 10 which is d with key F. so fail
So no choice 431 + "a" , 431+161 all in string
431a, 431161, 43197
or if  recrouse a will be
a431   stand chopping png
61431  walk chopping png
141431 jump chopping png
97431 gravity chopping png

if recouse b +walkman will be
a420   mining png
61420  samsheing png
141420  jumping smash png
97420   gravity smash png

Which if person b call  resourse a will be 420a, 42061, 420141, 42097 again will fail

Column1 Hexaldecimal octal Decimal Binary Bota function Bitwise key ^ f  with resource a with resource b with resource c with resource d with resource e with resource f with resource g
[ 5B 133 91 1011011 Ww== 2073613 [2073613
5B2073613
1332073613
912073613
           
\ need \\ sowon't use 5C 134 92 1011100 XA== 2073810 dont use            
] 5D 135 93 1011101 XQ== 2073811 ]2073811
5D2073811
1352073811
932073811
           
^ 5E 136 94 1011110 Xg== 2073832 ^2073832
5E2073832
1362073832
942073832
           
_ 5F 137 95 1011111 Xw== 2073833 _2073833
5F2073833
1372073833
952073833
           
a 61 141 97 1100001 YQ==   431 a431
61431
141431
97431
a420
61420
141420
97420
a421
61421
141421
97421
a10
6110
14110
9710
a11
6111
14111
9711
a0
610
1410
970
a1
611
1411
971
b 62 142 98 1100010 Yg== 420 b420
62420
142420
98420
b431
62431
142431
98431
b420
62420
142420
98420
       
c 63 143 99 1100011 Yw== 421 c421
63421
143421
99421
           
d 64 144 100 1100100 ZA== 10 d10
6410
14410
10010
           
e 65 145 101 1100101 ZQ== 11 e11
6511
14511
10111
           
f 66 146 102 1100110 Zg== 0 f0
660
1460
1020
           
g 67 147 103 1100111 Zw== 1 g1
671
1471
1031
           
h 68 150 104 1101000 aA== 1414 h1414
681414
1501414
1041414
       
i 69 151 105 1101001 aQ== 1415 i1415
691415
1511414
1051415
           
j 6A 152 106 1101010 ag== 1436 j1436
6A1436
1521436
1061436
           
k 6B 153 107 1101011 aw== 1437 k1437
6B1437
1531437
1071437
           
l 6C 154 108 1101100 bA== 1122 l1122
6C1122
1541122
1081122
           
m 6D 155 109 1101101 bQ== 1123 m1123
6D1123
1551123
1091123
           
n 6E 156 110 1101110 bg== 1144 n1144
6E1144
1561144
1101144
           
o 6F 157 111 1101111 bw== 1145 o1145
6F1145
1571145
1111145
           
p 70 160 112 1110000 cA== 9918 p9918
709918
1609918
1129918
           
q 71 161 113 1110001 cQ== 9919 q9919
719919
1619919
1139919
           
r 72 162 114 1110010 cg== 9908 r9908
729908
1629908
1149908
           
s 73 163 115 1110011 cw== 9909 s9909
739909
1639909
1159909
           
t 74 164 116 1110100 dA== 14618 t14618
7414618
16414618
11614618
           
u 75 165 117 1110101 dQ== 14619 u14619
7514619
16514619
11714619
           
v 76 166 118 1110110 dg== 14608 v14608
7614608
16614608
11814608
           
w 77 167 119 1110111 dw== 14609 w14609
7714609
16714609
11914609
           
x 78 170 120 1111000 eA== 14998 x14998
7814998
17014998
12014998
           
y 79 171 121 1111001 eQ== 14999 y14999
7914999
17114999
12114999
           
z 7A 172 122 1111010 eg== 15020 z15020
7A15020
17215020
12215020
       
{ 7B 173 123 1111011 ew== 15021 {15021
7B152021
17315021
12315021
           
| 7C 174 124 1111100 fA== 15730 |15730
7C15730
17415730
12415730
           
} 7D 175 125 1111101 fQ== 15731 }15731
7D15731
17515731
12515731
           
! 21 41 33 100001 IQ== 1134575 !1134575
211134575
411134575
331134575
           
# 23 43 35 100011 Iw== 1134565 #1134565
231134565
431134565
351134565
           
$ 24 44 36 100100 JA== 1134154 $1134154
241134154
441134154
361134154
           
% 25 45 37 100101 JQ== 1134155 %1134155
251134155
451134155
371134155
           
& 26 46 38 100110 Jg== 1134144 &1134144
261134144
461134144
381134144
           
( 28 50 40 101000 KA== 1131462 (113141462
28113141462
501131462
401131462
           
* 2A 52 42 101010 Kg== 1131484 *1131484
2A1131484
521131484
421131484
           
/ 2F 57 47 101111 Lw== 1131449 / cannot use        
0 30 60 48 110000 MA== 1139966 01139966
301139966
601139966
481139966
           
PERSON A EACH RESOURCE HAS 4 UNIQUE PNG TO MAKE (STANDING, WALKING,JUMPING,DROP) SO WHEN STAND MINING IS 1 PIC, JUMP MINING 2ND DROP MINING 3RD PIC, WALK MINGING IS 4TH PIC
[ ] ^ _
a2073613
612073613
1412073613
972073613
a2073811
612073811
1412073811
972073811
a2073832
612073832
1412073832
972073832
a2073833
612073833
1412073833
972073833
a b c d
a431
61431
141431
97431
a420
61420
141420
97420
a421
61421
141421
97421
a10
6110
14110
9710
e f g h
a11
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9711
a0
610
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a1
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971
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m n o p
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a9918
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q r s t
a9919
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a9908
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a9909
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u v w x
a14619
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a14608
6114608
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a14609
6114609
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a14998
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y z { |
a14999
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a15020
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a15021
6115021
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} ! # $
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a1134575
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a1134565
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a1134154
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% & ( *
a1134155
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a1134144
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971134144
a1131462
611131462
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a1131484
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0
a1139966
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