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G = C10×Q8⋊2S3order 480 = 25·3·5

Direct product of C10 and Q8⋊2S3

Series: Derived Chief Lower central Upper central

 Derived series C1 — C12 — C10×Q8⋊2S3
 Chief series C1 — C3 — C6 — C12 — C60 — C5×D12 — C10×D12 — C10×Q8⋊2S3
 Lower central C3 — C6 — C12 — C10×Q8⋊2S3
 Upper central C1 — C2×C10 — C2×C20 — Q8×C10

Generators and relations for C10×Q82S3
G = < a,b,c,d,e | a10=b4=d3=e2=1, c2=b2, ab=ba, ac=ca, ad=da, ae=ea, cbc-1=ebe=b-1, bd=db, cd=dc, ece=b-1c, ede=d-1 >

Subgroups: 356 in 136 conjugacy classes, 66 normal (34 characteristic)
C1, C2, C2, C2, C3, C4, C4, C22, C22, C5, S3, C6, C6, C8, C2×C4, C2×C4, D4, Q8, Q8, C23, C10, C10, C10, C12, C12, D6, C2×C6, C15, C2×C8, SD16, C2×D4, C2×Q8, C20, C20, C2×C10, C2×C10, C3⋊C8, D12, D12, C2×C12, C2×C12, C3×Q8, C3×Q8, C22×S3, C5×S3, C30, C30, C2×SD16, C40, C2×C20, C2×C20, C5×D4, C5×Q8, C5×Q8, C22×C10, C2×C3⋊C8, Q82S3, C2×D12, C6×Q8, C60, C60, S3×C10, C2×C30, C2×C40, C5×SD16, D4×C10, Q8×C10, C2×Q82S3, C5×C3⋊C8, C5×D12, C5×D12, C2×C60, C2×C60, Q8×C15, Q8×C15, S3×C2×C10, C10×SD16, C10×C3⋊C8, C5×Q82S3, C10×D12, Q8×C30, C10×Q82S3
Quotients: C1, C2, C22, C5, S3, D4, C23, C10, D6, SD16, C2×D4, C2×C10, C3⋊D4, C22×S3, C5×S3, C2×SD16, C5×D4, C22×C10, Q82S3, C2×C3⋊D4, S3×C10, C5×SD16, D4×C10, C2×Q82S3, C5×C3⋊D4, S3×C2×C10, C10×SD16, C5×Q82S3, C10×C3⋊D4, C10×Q82S3

Smallest permutation representation of C10×Q82S3
On 240 points
Generators in S240
(1 2 3 4 5 6 7 8 9 10)(11 12 13 14 15 16 17 18 19 20)(21 22 23 24 25 26 27 28 29 30)(31 32 33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48 49 50)(51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70)(71 72 73 74 75 76 77 78 79 80)(81 82 83 84 85 86 87 88 89 90)(91 92 93 94 95 96 97 98 99 100)(101 102 103 104 105 106 107 108 109 110)(111 112 113 114 115 116 117 118 119 120)(121 122 123 124 125 126 127 128 129 130)(131 132 133 134 135 136 137 138 139 140)(141 142 143 144 145 146 147 148 149 150)(151 152 153 154 155 156 157 158 159 160)(161 162 163 164 165 166 167 168 169 170)(171 172 173 174 175 176 177 178 179 180)(181 182 183 184 185 186 187 188 189 190)(191 192 193 194 195 196 197 198 199 200)(201 202 203 204 205 206 207 208 209 210)(211 212 213 214 215 216 217 218 219 220)(221 222 223 224 225 226 227 228 229 230)(231 232 233 234 235 236 237 238 239 240)
(1 197 38 185)(2 198 39 186)(3 199 40 187)(4 200 31 188)(5 191 32 189)(6 192 33 190)(7 193 34 181)(8 194 35 182)(9 195 36 183)(10 196 37 184)(11 159 206 116)(12 160 207 117)(13 151 208 118)(14 152 209 119)(15 153 210 120)(16 154 201 111)(17 155 202 112)(18 156 203 113)(19 157 204 114)(20 158 205 115)(21 230 78 123)(22 221 79 124)(23 222 80 125)(24 223 71 126)(25 224 72 127)(26 225 73 128)(27 226 74 129)(28 227 75 130)(29 228 76 121)(30 229 77 122)(41 141 98 55)(42 142 99 56)(43 143 100 57)(44 144 91 58)(45 145 92 59)(46 146 93 60)(47 147 94 51)(48 148 95 52)(49 149 96 53)(50 150 97 54)(61 136 239 101)(62 137 240 102)(63 138 231 103)(64 139 232 104)(65 140 233 105)(66 131 234 106)(67 132 235 107)(68 133 236 108)(69 134 237 109)(70 135 238 110)(81 162 217 177)(82 163 218 178)(83 164 219 179)(84 165 220 180)(85 166 211 171)(86 167 212 172)(87 168 213 173)(88 169 214 174)(89 170 215 175)(90 161 216 176)
(1 127 38 224)(2 128 39 225)(3 129 40 226)(4 130 31 227)(5 121 32 228)(6 122 33 229)(7 123 34 230)(8 124 35 221)(9 125 36 222)(10 126 37 223)(11 162 206 177)(12 163 207 178)(13 164 208 179)(14 165 209 180)(15 166 210 171)(16 167 201 172)(17 168 202 173)(18 169 203 174)(19 170 204 175)(20 161 205 176)(21 193 78 181)(22 194 79 182)(23 195 80 183)(24 196 71 184)(25 197 72 185)(26 198 73 186)(27 199 74 187)(28 200 75 188)(29 191 76 189)(30 192 77 190)(41 237 98 69)(42 238 99 70)(43 239 100 61)(44 240 91 62)(45 231 92 63)(46 232 93 64)(47 233 94 65)(48 234 95 66)(49 235 96 67)(50 236 97 68)(51 105 147 140)(52 106 148 131)(53 107 149 132)(54 108 150 133)(55 109 141 134)(56 110 142 135)(57 101 143 136)(58 102 144 137)(59 103 145 138)(60 104 146 139)(81 116 217 159)(82 117 218 160)(83 118 219 151)(84 119 220 152)(85 120 211 153)(86 111 212 154)(87 112 213 155)(88 113 214 156)(89 114 215 157)(90 115 216 158)
(1 61 162)(2 62 163)(3 63 164)(4 64 165)(5 65 166)(6 66 167)(7 67 168)(8 68 169)(9 69 170)(10 70 161)(11 224 100)(12 225 91)(13 226 92)(14 227 93)(15 228 94)(16 229 95)(17 230 96)(18 221 97)(19 222 98)(20 223 99)(21 149 112)(22 150 113)(23 141 114)(24 142 115)(25 143 116)(26 144 117)(27 145 118)(28 146 119)(29 147 120)(30 148 111)(31 232 180)(32 233 171)(33 234 172)(34 235 173)(35 236 174)(36 237 175)(37 238 176)(38 239 177)(39 240 178)(40 231 179)(41 204 125)(42 205 126)(43 206 127)(44 207 128)(45 208 129)(46 209 130)(47 210 121)(48 201 122)(49 202 123)(50 203 124)(51 153 76)(52 154 77)(53 155 78)(54 156 79)(55 157 80)(56 158 71)(57 159 72)(58 160 73)(59 151 74)(60 152 75)(81 185 101)(82 186 102)(83 187 103)(84 188 104)(85 189 105)(86 190 106)(87 181 107)(88 182 108)(89 183 109)(90 184 110)(131 212 192)(132 213 193)(133 214 194)(134 215 195)(135 216 196)(136 217 197)(137 218 198)(138 219 199)(139 220 200)(140 211 191)
(1 6)(2 7)(3 8)(4 9)(5 10)(11 52)(12 53)(13 54)(14 55)(15 56)(16 57)(17 58)(18 59)(19 60)(20 51)(21 128)(22 129)(23 130)(24 121)(25 122)(26 123)(27 124)(28 125)(29 126)(30 127)(31 36)(32 37)(33 38)(34 39)(35 40)(41 119)(42 120)(43 111)(44 112)(45 113)(46 114)(47 115)(48 116)(49 117)(50 118)(61 167)(62 168)(63 169)(64 170)(65 161)(66 162)(67 163)(68 164)(69 165)(70 166)(71 228)(72 229)(73 230)(74 221)(75 222)(76 223)(77 224)(78 225)(79 226)(80 227)(81 131)(82 132)(83 133)(84 134)(85 135)(86 136)(87 137)(88 138)(89 139)(90 140)(91 155)(92 156)(93 157)(94 158)(95 159)(96 160)(97 151)(98 152)(99 153)(100 154)(101 212)(102 213)(103 214)(104 215)(105 216)(106 217)(107 218)(108 219)(109 220)(110 211)(141 209)(142 210)(143 201)(144 202)(145 203)(146 204)(147 205)(148 206)(149 207)(150 208)(171 238)(172 239)(173 240)(174 231)(175 232)(176 233)(177 234)(178 235)(179 236)(180 237)(181 198)(182 199)(183 200)(184 191)(185 192)(186 193)(187 194)(188 195)(189 196)(190 197)

G:=sub<Sym(240)| (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50)(51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110)(111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130)(131,132,133,134,135,136,137,138,139,140)(141,142,143,144,145,146,147,148,149,150)(151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170)(171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190)(191,192,193,194,195,196,197,198,199,200)(201,202,203,204,205,206,207,208,209,210)(211,212,213,214,215,216,217,218,219,220)(221,222,223,224,225,226,227,228,229,230)(231,232,233,234,235,236,237,238,239,240), (1,197,38,185)(2,198,39,186)(3,199,40,187)(4,200,31,188)(5,191,32,189)(6,192,33,190)(7,193,34,181)(8,194,35,182)(9,195,36,183)(10,196,37,184)(11,159,206,116)(12,160,207,117)(13,151,208,118)(14,152,209,119)(15,153,210,120)(16,154,201,111)(17,155,202,112)(18,156,203,113)(19,157,204,114)(20,158,205,115)(21,230,78,123)(22,221,79,124)(23,222,80,125)(24,223,71,126)(25,224,72,127)(26,225,73,128)(27,226,74,129)(28,227,75,130)(29,228,76,121)(30,229,77,122)(41,141,98,55)(42,142,99,56)(43,143,100,57)(44,144,91,58)(45,145,92,59)(46,146,93,60)(47,147,94,51)(48,148,95,52)(49,149,96,53)(50,150,97,54)(61,136,239,101)(62,137,240,102)(63,138,231,103)(64,139,232,104)(65,140,233,105)(66,131,234,106)(67,132,235,107)(68,133,236,108)(69,134,237,109)(70,135,238,110)(81,162,217,177)(82,163,218,178)(83,164,219,179)(84,165,220,180)(85,166,211,171)(86,167,212,172)(87,168,213,173)(88,169,214,174)(89,170,215,175)(90,161,216,176), (1,127,38,224)(2,128,39,225)(3,129,40,226)(4,130,31,227)(5,121,32,228)(6,122,33,229)(7,123,34,230)(8,124,35,221)(9,125,36,222)(10,126,37,223)(11,162,206,177)(12,163,207,178)(13,164,208,179)(14,165,209,180)(15,166,210,171)(16,167,201,172)(17,168,202,173)(18,169,203,174)(19,170,204,175)(20,161,205,176)(21,193,78,181)(22,194,79,182)(23,195,80,183)(24,196,71,184)(25,197,72,185)(26,198,73,186)(27,199,74,187)(28,200,75,188)(29,191,76,189)(30,192,77,190)(41,237,98,69)(42,238,99,70)(43,239,100,61)(44,240,91,62)(45,231,92,63)(46,232,93,64)(47,233,94,65)(48,234,95,66)(49,235,96,67)(50,236,97,68)(51,105,147,140)(52,106,148,131)(53,107,149,132)(54,108,150,133)(55,109,141,134)(56,110,142,135)(57,101,143,136)(58,102,144,137)(59,103,145,138)(60,104,146,139)(81,116,217,159)(82,117,218,160)(83,118,219,151)(84,119,220,152)(85,120,211,153)(86,111,212,154)(87,112,213,155)(88,113,214,156)(89,114,215,157)(90,115,216,158), (1,61,162)(2,62,163)(3,63,164)(4,64,165)(5,65,166)(6,66,167)(7,67,168)(8,68,169)(9,69,170)(10,70,161)(11,224,100)(12,225,91)(13,226,92)(14,227,93)(15,228,94)(16,229,95)(17,230,96)(18,221,97)(19,222,98)(20,223,99)(21,149,112)(22,150,113)(23,141,114)(24,142,115)(25,143,116)(26,144,117)(27,145,118)(28,146,119)(29,147,120)(30,148,111)(31,232,180)(32,233,171)(33,234,172)(34,235,173)(35,236,174)(36,237,175)(37,238,176)(38,239,177)(39,240,178)(40,231,179)(41,204,125)(42,205,126)(43,206,127)(44,207,128)(45,208,129)(46,209,130)(47,210,121)(48,201,122)(49,202,123)(50,203,124)(51,153,76)(52,154,77)(53,155,78)(54,156,79)(55,157,80)(56,158,71)(57,159,72)(58,160,73)(59,151,74)(60,152,75)(81,185,101)(82,186,102)(83,187,103)(84,188,104)(85,189,105)(86,190,106)(87,181,107)(88,182,108)(89,183,109)(90,184,110)(131,212,192)(132,213,193)(133,214,194)(134,215,195)(135,216,196)(136,217,197)(137,218,198)(138,219,199)(139,220,200)(140,211,191), (1,6)(2,7)(3,8)(4,9)(5,10)(11,52)(12,53)(13,54)(14,55)(15,56)(16,57)(17,58)(18,59)(19,60)(20,51)(21,128)(22,129)(23,130)(24,121)(25,122)(26,123)(27,124)(28,125)(29,126)(30,127)(31,36)(32,37)(33,38)(34,39)(35,40)(41,119)(42,120)(43,111)(44,112)(45,113)(46,114)(47,115)(48,116)(49,117)(50,118)(61,167)(62,168)(63,169)(64,170)(65,161)(66,162)(67,163)(68,164)(69,165)(70,166)(71,228)(72,229)(73,230)(74,221)(75,222)(76,223)(77,224)(78,225)(79,226)(80,227)(81,131)(82,132)(83,133)(84,134)(85,135)(86,136)(87,137)(88,138)(89,139)(90,140)(91,155)(92,156)(93,157)(94,158)(95,159)(96,160)(97,151)(98,152)(99,153)(100,154)(101,212)(102,213)(103,214)(104,215)(105,216)(106,217)(107,218)(108,219)(109,220)(110,211)(141,209)(142,210)(143,201)(144,202)(145,203)(146,204)(147,205)(148,206)(149,207)(150,208)(171,238)(172,239)(173,240)(174,231)(175,232)(176,233)(177,234)(178,235)(179,236)(180,237)(181,198)(182,199)(183,200)(184,191)(185,192)(186,193)(187,194)(188,195)(189,196)(190,197)>;

G:=Group( (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50)(51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110)(111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130)(131,132,133,134,135,136,137,138,139,140)(141,142,143,144,145,146,147,148,149,150)(151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170)(171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190)(191,192,193,194,195,196,197,198,199,200)(201,202,203,204,205,206,207,208,209,210)(211,212,213,214,215,216,217,218,219,220)(221,222,223,224,225,226,227,228,229,230)(231,232,233,234,235,236,237,238,239,240), (1,197,38,185)(2,198,39,186)(3,199,40,187)(4,200,31,188)(5,191,32,189)(6,192,33,190)(7,193,34,181)(8,194,35,182)(9,195,36,183)(10,196,37,184)(11,159,206,116)(12,160,207,117)(13,151,208,118)(14,152,209,119)(15,153,210,120)(16,154,201,111)(17,155,202,112)(18,156,203,113)(19,157,204,114)(20,158,205,115)(21,230,78,123)(22,221,79,124)(23,222,80,125)(24,223,71,126)(25,224,72,127)(26,225,73,128)(27,226,74,129)(28,227,75,130)(29,228,76,121)(30,229,77,122)(41,141,98,55)(42,142,99,56)(43,143,100,57)(44,144,91,58)(45,145,92,59)(46,146,93,60)(47,147,94,51)(48,148,95,52)(49,149,96,53)(50,150,97,54)(61,136,239,101)(62,137,240,102)(63,138,231,103)(64,139,232,104)(65,140,233,105)(66,131,234,106)(67,132,235,107)(68,133,236,108)(69,134,237,109)(70,135,238,110)(81,162,217,177)(82,163,218,178)(83,164,219,179)(84,165,220,180)(85,166,211,171)(86,167,212,172)(87,168,213,173)(88,169,214,174)(89,170,215,175)(90,161,216,176), (1,127,38,224)(2,128,39,225)(3,129,40,226)(4,130,31,227)(5,121,32,228)(6,122,33,229)(7,123,34,230)(8,124,35,221)(9,125,36,222)(10,126,37,223)(11,162,206,177)(12,163,207,178)(13,164,208,179)(14,165,209,180)(15,166,210,171)(16,167,201,172)(17,168,202,173)(18,169,203,174)(19,170,204,175)(20,161,205,176)(21,193,78,181)(22,194,79,182)(23,195,80,183)(24,196,71,184)(25,197,72,185)(26,198,73,186)(27,199,74,187)(28,200,75,188)(29,191,76,189)(30,192,77,190)(41,237,98,69)(42,238,99,70)(43,239,100,61)(44,240,91,62)(45,231,92,63)(46,232,93,64)(47,233,94,65)(48,234,95,66)(49,235,96,67)(50,236,97,68)(51,105,147,140)(52,106,148,131)(53,107,149,132)(54,108,150,133)(55,109,141,134)(56,110,142,135)(57,101,143,136)(58,102,144,137)(59,103,145,138)(60,104,146,139)(81,116,217,159)(82,117,218,160)(83,118,219,151)(84,119,220,152)(85,120,211,153)(86,111,212,154)(87,112,213,155)(88,113,214,156)(89,114,215,157)(90,115,216,158), (1,61,162)(2,62,163)(3,63,164)(4,64,165)(5,65,166)(6,66,167)(7,67,168)(8,68,169)(9,69,170)(10,70,161)(11,224,100)(12,225,91)(13,226,92)(14,227,93)(15,228,94)(16,229,95)(17,230,96)(18,221,97)(19,222,98)(20,223,99)(21,149,112)(22,150,113)(23,141,114)(24,142,115)(25,143,116)(26,144,117)(27,145,118)(28,146,119)(29,147,120)(30,148,111)(31,232,180)(32,233,171)(33,234,172)(34,235,173)(35,236,174)(36,237,175)(37,238,176)(38,239,177)(39,240,178)(40,231,179)(41,204,125)(42,205,126)(43,206,127)(44,207,128)(45,208,129)(46,209,130)(47,210,121)(48,201,122)(49,202,123)(50,203,124)(51,153,76)(52,154,77)(53,155,78)(54,156,79)(55,157,80)(56,158,71)(57,159,72)(58,160,73)(59,151,74)(60,152,75)(81,185,101)(82,186,102)(83,187,103)(84,188,104)(85,189,105)(86,190,106)(87,181,107)(88,182,108)(89,183,109)(90,184,110)(131,212,192)(132,213,193)(133,214,194)(134,215,195)(135,216,196)(136,217,197)(137,218,198)(138,219,199)(139,220,200)(140,211,191), (1,6)(2,7)(3,8)(4,9)(5,10)(11,52)(12,53)(13,54)(14,55)(15,56)(16,57)(17,58)(18,59)(19,60)(20,51)(21,128)(22,129)(23,130)(24,121)(25,122)(26,123)(27,124)(28,125)(29,126)(30,127)(31,36)(32,37)(33,38)(34,39)(35,40)(41,119)(42,120)(43,111)(44,112)(45,113)(46,114)(47,115)(48,116)(49,117)(50,118)(61,167)(62,168)(63,169)(64,170)(65,161)(66,162)(67,163)(68,164)(69,165)(70,166)(71,228)(72,229)(73,230)(74,221)(75,222)(76,223)(77,224)(78,225)(79,226)(80,227)(81,131)(82,132)(83,133)(84,134)(85,135)(86,136)(87,137)(88,138)(89,139)(90,140)(91,155)(92,156)(93,157)(94,158)(95,159)(96,160)(97,151)(98,152)(99,153)(100,154)(101,212)(102,213)(103,214)(104,215)(105,216)(106,217)(107,218)(108,219)(109,220)(110,211)(141,209)(142,210)(143,201)(144,202)(145,203)(146,204)(147,205)(148,206)(149,207)(150,208)(171,238)(172,239)(173,240)(174,231)(175,232)(176,233)(177,234)(178,235)(179,236)(180,237)(181,198)(182,199)(183,200)(184,191)(185,192)(186,193)(187,194)(188,195)(189,196)(190,197) );

G=PermutationGroup([[(1,2,3,4,5,6,7,8,9,10),(11,12,13,14,15,16,17,18,19,20),(21,22,23,24,25,26,27,28,29,30),(31,32,33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48,49,50),(51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70),(71,72,73,74,75,76,77,78,79,80),(81,82,83,84,85,86,87,88,89,90),(91,92,93,94,95,96,97,98,99,100),(101,102,103,104,105,106,107,108,109,110),(111,112,113,114,115,116,117,118,119,120),(121,122,123,124,125,126,127,128,129,130),(131,132,133,134,135,136,137,138,139,140),(141,142,143,144,145,146,147,148,149,150),(151,152,153,154,155,156,157,158,159,160),(161,162,163,164,165,166,167,168,169,170),(171,172,173,174,175,176,177,178,179,180),(181,182,183,184,185,186,187,188,189,190),(191,192,193,194,195,196,197,198,199,200),(201,202,203,204,205,206,207,208,209,210),(211,212,213,214,215,216,217,218,219,220),(221,222,223,224,225,226,227,228,229,230),(231,232,233,234,235,236,237,238,239,240)], [(1,197,38,185),(2,198,39,186),(3,199,40,187),(4,200,31,188),(5,191,32,189),(6,192,33,190),(7,193,34,181),(8,194,35,182),(9,195,36,183),(10,196,37,184),(11,159,206,116),(12,160,207,117),(13,151,208,118),(14,152,209,119),(15,153,210,120),(16,154,201,111),(17,155,202,112),(18,156,203,113),(19,157,204,114),(20,158,205,115),(21,230,78,123),(22,221,79,124),(23,222,80,125),(24,223,71,126),(25,224,72,127),(26,225,73,128),(27,226,74,129),(28,227,75,130),(29,228,76,121),(30,229,77,122),(41,141,98,55),(42,142,99,56),(43,143,100,57),(44,144,91,58),(45,145,92,59),(46,146,93,60),(47,147,94,51),(48,148,95,52),(49,149,96,53),(50,150,97,54),(61,136,239,101),(62,137,240,102),(63,138,231,103),(64,139,232,104),(65,140,233,105),(66,131,234,106),(67,132,235,107),(68,133,236,108),(69,134,237,109),(70,135,238,110),(81,162,217,177),(82,163,218,178),(83,164,219,179),(84,165,220,180),(85,166,211,171),(86,167,212,172),(87,168,213,173),(88,169,214,174),(89,170,215,175),(90,161,216,176)], [(1,127,38,224),(2,128,39,225),(3,129,40,226),(4,130,31,227),(5,121,32,228),(6,122,33,229),(7,123,34,230),(8,124,35,221),(9,125,36,222),(10,126,37,223),(11,162,206,177),(12,163,207,178),(13,164,208,179),(14,165,209,180),(15,166,210,171),(16,167,201,172),(17,168,202,173),(18,169,203,174),(19,170,204,175),(20,161,205,176),(21,193,78,181),(22,194,79,182),(23,195,80,183),(24,196,71,184),(25,197,72,185),(26,198,73,186),(27,199,74,187),(28,200,75,188),(29,191,76,189),(30,192,77,190),(41,237,98,69),(42,238,99,70),(43,239,100,61),(44,240,91,62),(45,231,92,63),(46,232,93,64),(47,233,94,65),(48,234,95,66),(49,235,96,67),(50,236,97,68),(51,105,147,140),(52,106,148,131),(53,107,149,132),(54,108,150,133),(55,109,141,134),(56,110,142,135),(57,101,143,136),(58,102,144,137),(59,103,145,138),(60,104,146,139),(81,116,217,159),(82,117,218,160),(83,118,219,151),(84,119,220,152),(85,120,211,153),(86,111,212,154),(87,112,213,155),(88,113,214,156),(89,114,215,157),(90,115,216,158)], [(1,61,162),(2,62,163),(3,63,164),(4,64,165),(5,65,166),(6,66,167),(7,67,168),(8,68,169),(9,69,170),(10,70,161),(11,224,100),(12,225,91),(13,226,92),(14,227,93),(15,228,94),(16,229,95),(17,230,96),(18,221,97),(19,222,98),(20,223,99),(21,149,112),(22,150,113),(23,141,114),(24,142,115),(25,143,116),(26,144,117),(27,145,118),(28,146,119),(29,147,120),(30,148,111),(31,232,180),(32,233,171),(33,234,172),(34,235,173),(35,236,174),(36,237,175),(37,238,176),(38,239,177),(39,240,178),(40,231,179),(41,204,125),(42,205,126),(43,206,127),(44,207,128),(45,208,129),(46,209,130),(47,210,121),(48,201,122),(49,202,123),(50,203,124),(51,153,76),(52,154,77),(53,155,78),(54,156,79),(55,157,80),(56,158,71),(57,159,72),(58,160,73),(59,151,74),(60,152,75),(81,185,101),(82,186,102),(83,187,103),(84,188,104),(85,189,105),(86,190,106),(87,181,107),(88,182,108),(89,183,109),(90,184,110),(131,212,192),(132,213,193),(133,214,194),(134,215,195),(135,216,196),(136,217,197),(137,218,198),(138,219,199),(139,220,200),(140,211,191)], [(1,6),(2,7),(3,8),(4,9),(5,10),(11,52),(12,53),(13,54),(14,55),(15,56),(16,57),(17,58),(18,59),(19,60),(20,51),(21,128),(22,129),(23,130),(24,121),(25,122),(26,123),(27,124),(28,125),(29,126),(30,127),(31,36),(32,37),(33,38),(34,39),(35,40),(41,119),(42,120),(43,111),(44,112),(45,113),(46,114),(47,115),(48,116),(49,117),(50,118),(61,167),(62,168),(63,169),(64,170),(65,161),(66,162),(67,163),(68,164),(69,165),(70,166),(71,228),(72,229),(73,230),(74,221),(75,222),(76,223),(77,224),(78,225),(79,226),(80,227),(81,131),(82,132),(83,133),(84,134),(85,135),(86,136),(87,137),(88,138),(89,139),(90,140),(91,155),(92,156),(93,157),(94,158),(95,159),(96,160),(97,151),(98,152),(99,153),(100,154),(101,212),(102,213),(103,214),(104,215),(105,216),(106,217),(107,218),(108,219),(109,220),(110,211),(141,209),(142,210),(143,201),(144,202),(145,203),(146,204),(147,205),(148,206),(149,207),(150,208),(171,238),(172,239),(173,240),(174,231),(175,232),(176,233),(177,234),(178,235),(179,236),(180,237),(181,198),(182,199),(183,200),(184,191),(185,192),(186,193),(187,194),(188,195),(189,196),(190,197)]])

120 conjugacy classes

 class 1 2A 2B 2C 2D 2E 3 4A 4B 4C 4D 5A 5B 5C 5D 6A 6B 6C 8A 8B 8C 8D 10A ··· 10L 10M ··· 10T 12A ··· 12F 15A 15B 15C 15D 20A ··· 20H 20I ··· 20P 30A ··· 30L 40A ··· 40P 60A ··· 60X order 1 2 2 2 2 2 3 4 4 4 4 5 5 5 5 6 6 6 8 8 8 8 10 ··· 10 10 ··· 10 12 ··· 12 15 15 15 15 20 ··· 20 20 ··· 20 30 ··· 30 40 ··· 40 60 ··· 60 size 1 1 1 1 12 12 2 2 2 4 4 1 1 1 1 2 2 2 6 6 6 6 1 ··· 1 12 ··· 12 4 ··· 4 2 2 2 2 2 ··· 2 4 ··· 4 2 ··· 2 6 ··· 6 4 ··· 4

120 irreducible representations

 dim 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 type + + + + + + + + + + + image C1 C2 C2 C2 C2 C5 C10 C10 C10 C10 S3 D4 D4 D6 D6 SD16 C3⋊D4 C3⋊D4 C5×S3 C5×D4 C5×D4 S3×C10 S3×C10 C5×SD16 C5×C3⋊D4 C5×C3⋊D4 Q8⋊2S3 C5×Q8⋊2S3 kernel C10×Q8⋊2S3 C10×C3⋊C8 C5×Q8⋊2S3 C10×D12 Q8×C30 C2×Q8⋊2S3 C2×C3⋊C8 Q8⋊2S3 C2×D12 C6×Q8 Q8×C10 C60 C2×C30 C2×C20 C5×Q8 C30 C20 C2×C10 C2×Q8 C12 C2×C6 C2×C4 Q8 C6 C4 C22 C10 C2 # reps 1 1 4 1 1 4 4 16 4 4 1 1 1 1 2 4 2 2 4 4 4 4 8 16 8 8 2 8

Matrix representation of C10×Q82S3 in GL4(𝔽241) generated by

 150 0 0 0 0 150 0 0 0 0 154 0 0 0 0 154
,
 240 0 0 0 0 240 0 0 0 0 240 3 0 0 160 1
,
 70 140 0 0 101 171 0 0 0 0 0 57 0 0 93 0
,
 0 1 0 0 240 240 0 0 0 0 1 0 0 0 0 1
,
 0 1 0 0 1 0 0 0 0 0 240 0 0 0 160 1
G:=sub<GL(4,GF(241))| [150,0,0,0,0,150,0,0,0,0,154,0,0,0,0,154],[240,0,0,0,0,240,0,0,0,0,240,160,0,0,3,1],[70,101,0,0,140,171,0,0,0,0,0,93,0,0,57,0],[0,240,0,0,1,240,0,0,0,0,1,0,0,0,0,1],[0,1,0,0,1,0,0,0,0,0,240,160,0,0,0,1] >;

C10×Q82S3 in GAP, Magma, Sage, TeX

C_{10}\times Q_8\rtimes_2S_3
% in TeX

G:=Group("C10xQ8:2S3");
// GroupNames label

G:=SmallGroup(480,820);
// by ID

G=gap.SmallGroup(480,820);
# by ID

G:=PCGroup([7,-2,-2,-2,-5,-2,-2,-3,926,436,4204,1068,102,15686]);
// Polycyclic

G:=Group<a,b,c,d,e|a^10=b^4=d^3=e^2=1,c^2=b^2,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,c*b*c^-1=e*b*e=b^-1,b*d=d*b,c*d=d*c,e*c*e=b^-1*c,e*d*e=d^-1>;
// generators/relations

׿
×
𝔽


text:
# How to show, this is not vector space

The set of all vectors $V = [v_1 , v_2 , v_3 , v_4]$ $R^4$ with $v_1+v_2=0$ and $v_3-v_4=1$ is not a vector space.

format: plain


text:
## anonymous 3 years ago Dividing fractions..

1. KingGeorge

So suppose you have a fraction $\Large \frac{5/4}{7/3}$Then this can be simplified by doing $\Large \frac{5}{4}\times\frac{3}{7}=\frac{15}{28}$

2. anonymous

Division is defined as multiplication by the reciprocal, yes.

3. KingGeorge

For another example, if you have $\Large \frac{5/4}{7}$Then this is the same thing as $\Large \frac{5/4}{7/1}=\frac{5}{7}\times \frac{1}{4}=\frac{5}{28}$

4. anonymous

Interestingly enough, $$\large 5/4 ÷ 7/3 = \frac{5÷7}{4÷3} = 5/7 ÷ 4/3.$$

5. KingGeorge

Now using the examples I just gave, can you tell me what $\Large \frac{3/2}{6/5}$can be simplified to?

6. anonymous

In general we can say when you divide, a$(a \div b)\div(c \div d) = (a \div b)\times(d \div c)$

7. anonymous

Is it.. 1 1/4

8. KingGeorge

That's correct.

9. KingGeorge

Now, let's mix it up a bit. What about $\Large \frac{8}{2/5}$If this is a bit confusing, try turning the 8 into a fraction and seeing what you can do then.

10. anonymous

Um.. 20?

11. KingGeorge

Right again. I think you're getting the hang of this. Does this help you with the problems you were given?

12. anonymous

Yeah, it does.

13. anonymous

Thanks.

14. KingGeorge

You're welcome.

15. KingGeorge

I strongly believe in the power of varying examples that are usually more difficult than necessary.


text:
# Tag Info

107

The difference between the PKCS#5 and PKCS#7 padding mechanisms is the block size; PKCS#5 padding is defined for 8-byte block sizes, PKCS#7 padding would work for any block size from 1 to 255 bytes. This is the definition of PKCS#5 padding (6.2) as defined in the RFC: The padding string PS shall consist of 8 - (||M|| mod 8) octets all having value 8 - (...

46

Short: CBC mode in context of TLS protocol has had security issues, and would have had to be reworked. AES-CBC mode combined with decent HMAC can be as secure as AES-GCM. However, combining the cipher and MAC securely has been in practice found to be much easier said than done. Also, padding that is required by AES-CBC mode complicates things. In ...

37

XTS vs. Undiffused CBC. The issue here is malleability. Both XTS and CBC prevent an attacker from learning information about encrypted data. However, neither one completely succeeds in preventing an attacker from tampering with encrypted data. However, it's arguably easier to tamper with an (undiffused) CBC ciphertext than it is to tamper with an XTS ...

32

In comparison to CBC mode and HMAC, GCM mode is quite a commonly better alternative. But, I'll go to detail where it necessarily is not. Just like Richie Frame, I also do not agree that CBC + HMAC is always the best comparison target. I've added a few other details. Hope you find them useful. Against CBC and HMAC I'll discuss downsides first. The ...

31

The crucial difference between plain encryption and authenticated encryption (AE) is that AE additionally provides authenticity, while plain encryption provides only confidentiality. Let's investigate in detail these two notions. In the further text, we assume $K$ to be a secret key, which is known to authorized parties, but unknown to attackers. Goals ...

23

Contrary to what Stephen says, you absolutely can compute the tag in parallel. Here's how it works; the tag computation is essentially "assemble the AAD, data, the length field and $Encr(Nonce)$ into a series of values $x_n, x_{n-1}, x_{n-2}, ..., x_0$", and then "compute the polynomial $x_nh^n + x_{n-1}h^{n-1} + x_{n-2}h^{n-2} + ... + x_0h^0$ This ...

21

CBC does not perform authentication This property makes it less suitable for places where authentication is required, basically any transport protocol. TLS uses CBC, but by default performs authentication over the plain text instead of the ciphertext, which opened up a host of attacks. CBC can be used here, but it is error prone and may require an ...

19

There are some serious problems with this design that would preclude it from being standardized, so it probably does not have a name. The 2 visibly main flaws are as follows: If the plaintext follows a pattern similar to the block counter, the block cipher inputs may repeat, exposing information about the plaintext (exact same issue as reuse of nonce, but ...

17

The security of that approach is equivalent to that of normal CBC. Your scheme with first plaintext block $IV^\prime$ is clearly identical to normal CBC with $IV=AES(IV^\prime)$. Since a block cipher is a permutation over a block, a uniformly random first plaintext block will lead to a uniformly random IV for normal CBC. A ciphertext produced with your ...

17

A block cipher is an invertible transformation that maps an $n$ bit block of bits to an $n$ bit block of bits, under the control of a key (and where $n=128$ in the case of AES) Now, we most often need to do things other than mapping blocks of $n$ bits; how we do that is using the block cipher within a Mode of Operation. A mode of operation is just a way to ...

17

TLS 1.3 is a reboot of the TLS protocol which focused on up to date cryptography rather than backwards compatibility. Now CBC is not as secure as you make it to be, and the way that it was used in TLS made it particularly vulnerable. To note: in TLS the HMAC authentication tag was created over the plaintext rather than the ciphertext. This made TLS ...

15

The algorithm (now reasonably clear) is reminiscent of a block cipher in CFB mode, with $random$ as the IV (which can be public), $secret$ as the key, and MD5 used as keystream generator instead of the block cipher. Decryption works as in CFB: \begin{align*} M_1 &= C_1 \oplus \operatorname{MD5}( secret\mathbin\|random )\\ M_n &= C_n \oplus \...

15

Do not use a fixed IV. It can have seriously negative consequences. This is especially true for CBC mode. That said, a random 128-bit IV stored in plaintext is typically what you want. The IV can be known to an attacker without breaking security.

14

Thomas is correct; there's no attack on CFB mode if you can predict the IV; NIST is just being cautious. With CBC, the value of the first encrypted block $C_0 = E_k( IV \oplus P_0)$, where $IV$ is the IV used for that packet, $P_0$ is the value of the first plaintext block, and $E_k$ is the evaluation of the block cipher. If an attacker can predict the ...

14

Short answer: There would be nothing (that isn't already wrong with TLS) necessarily wrong with a CTR + HMAC cipher suite, but the technical merits are only one factor in a technical feature getting to RFC status in the TLS working group. Without being discourteous to the TLS Working Group (WG) participants or process, other reasons can be: political (...

13

While you do operate block-by-block when generating the pseudorandom stream, the actual encryption step (i.e., the XOR) is bitwise, and therefore does not require the message to be padded. For example, the message "Hello" will be processed as follows (pseudocode): byte stream[16] = AES(Key, Nonce); byte plaintext[5] = "Hello"; byte ciphertext[5]; for i ...

12

The MAC value should be calculated over all of the input, not just the first block. The chaining of CBC makes sure that the bits in the last block of ciphertext depends on all the previous blocks.

11

If you look at the CBC diagram, you'll see that having a fixed IV is equivalent to having the first ciphertext block become the IV. If your cipher is a good pseudorandom permutation, then what you are doing does work, if and only if all timestamps are unique such that the "new IV" is unique and unpredictable. And in fact, if you do not use the decrypted ...

11

An OCB like mode seems impossible with stream-ciphers. It's coupled tightly to the concept of a keyed permutation i.e. a (tweakable) block-cipher. Many authenticated encryption actually combine two distinct primitives. It's just that the specification and API only expose the combination. Essentially these xor a key-stream into the message to encrypt it (i....

11

With CBC mode the initialization vector is referred to as IV, because it is not nonce. There are ways to construct nonce so that it does not meet the needs of CBC mode. Random IV is one generation choice which is usually fine. Nonce can also be a counter, which is not ok here. Definitions Nonce means number used once. IV means initialization vector. CBC ...

11

The reference for this is NIST SP800-38A, especially its appendix B. Basically we consider the IV a binary value of the width of the block cipher (64-bit for DES, 128-bit for AES), and add 1 to that, except for one detail: there is no carry at some application-specified rank, defining the maximum number of blocks that can be enciphered with a single IV; if ...

11

The infinite garble extension makes sure that if a ciphertext block is changed that this block and each block after it doesn't decrypt correctly. The way that additional plaintext is affected when the ciphertext is changed is called error propagation. Error propagation over large parts of the plaintext is mainly interesting if you want to combine it with ...

10

An initialization vector never needs to be secret. If it needed to be secret then it would be called a "key" instead of a "vector". Generally, the initialization vector requirements are only dependent on the mode of operation, not the choice of the block cipher. The parameters of the mode of operation, such as the IV size, may however depend on the block ...

10

A mode of operation is an explicit method by which we use a block cipher (eg AES) to do more than just encrypt one block of data. For example, it may allow us to encrypt multiple blocks of data (eg ECB,CBC etc), provide us with some authenticated encryption (eg GCM) or a method for encrypting disc storage (eg XTS). Rijndael,DES etc are block ciphers. That ...

10

I do not remember if we checked this explicitly, but my guess is that in the chosen-plaintext setting the biclique attack would still be faster than the exhaustive search, maybe by the factor of 2 compared with 4 in the chosen-ciphertext setting. However, both results are pretty far from declaring AES broken in any sense. Such small gain over exhaustive ...

10

AES-GCM uses single block cipher operation and can be processed in parallel, therefore it should be faster. CTR+HMAC requires block cipher and hash function, which usually can't be processed in parallel. Also it requires 2 keys. It is often miss-implemented (MAC-then-encrypt or MAC-and-encrypt, using single key). Cipher-text length is the same for same ...

10

In CBC mode the decryption equation is $P_i = D_k(C_i) \oplus C_{i-1}$. If you received a corrupted $C_i$, $P_i$ and $P_{i+1}$ will be decrypted wrong, but $P_{i+2}$ no longer depends on $C_i$ and will be correct.

10

In the padding oracle attack you have an oracle that only tells you whether a particular chosen ciphertext decrypts to a correctly padded plaintext. That oracle is used to build a last word oracle, which used iteratively can reveal a whole message. The reason it works in CBC mode is that we can make predictable, arbitrary changes to the plaintext of the ...

10

You basically want a full disk encryption mode for a block cipher; XTS mode seems to be the current standard. In your case each "disk block" is actually a file offset. Note that using a stream cipher or counter mode is NOT secure if the data is ever modified in the file, as it would violate the cardinal sin of using the same key and initialization vector to ...

10

It is not secure, because an attacker can "mix and match" the output blocks from different authentication tags on different input messages, or repeat output blocks for repeated input blocks. For example, if the attacker knows the tag $F_k(m)$ for a one-block message $m$, then it can forge the correct tag $F_k(m) \mid F_k(m)$ for the two-block message \$m \...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# CO emissions in China: Uncertainties and implications of improved energy efficiency and emission control

 Title: CO emissions in China: Uncertainties and implications of improved energy efficiency and emission control Author: Zhao, Yu; Nielsen, Chris; McElroy, Michael Brendon; Zhang, Lin; Zhang, Jie Note: Order does not necessarily reflect citation order of authors. Citation: Zhao, Yu, Chris P. Nielsen, Michael B. McElroy, Lin Zhang, and Jie Zhang. 2012. “CO Emissions in China: Uncertainties and Implications of Improved Energy Efficiency and Emission Control.” Atmospheric Environment 49 (March): 103–113. doi:10.1016/j.atmosenv.2011.12.015. Access Status: Full text of the requested work is not available in DASH at this time (“dark deposit”). For more information on dark deposits, see our FAQ. Full Text & Related Files: Zhao et al. 1203 Atmos Envt CO Final.pdf (689.5Kb; PDF) Abstract: A bottom-up methodology and an improved database of emission factors combining the latest domestic field measurements are developed to estimate the emissions of anthropogenic CO from China at national and provincial levels. The CO emission factors for major economic sectors declined to varying degrees from 2005 to 2009, attributed to improved energy efficiency and/or emission control regulations. Total national CO emissions are estimated at 173 Tg for 2005 and have been relatively stable for subsequent years, despite fast growth of energy consumption and industrial production. While industry and transportation sources dominated CO emissions in developed eastern and north-central China, residential combustion played a much greater role in the less developed western provinces. The uncertainties of national Chinese CO emissions are quantified using Monte Carlo simulation at 20% to þ45% (95% confidence interval). Due to poor understanding of emission factors and activity levels for combustion of solid fuels, the largest uncertainties are found for emissions from the residential sector. The trends of bottom-up emissions compare reasonably to satellite observation of CO columns and to ground observations of CO2eCO correlation slopes. The increase in the ratio for emissions of CO2 relative to CO suggests that China has successfully improved combustion efficiencies across its economy in recent years, consistent with national policies to improve energy efficiency and to control criteria air pollutants. Published Version: doi:10.1016/j.atmosenv.2011.12.015 Citable link to this page: http://nrs.harvard.edu/urn-3:HUL.InstRepos:34730506 Downloads of this work:


text:
#### Archived

This topic is now archived and is closed to further replies.

# Lighting problems

This topic is 5742 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I'm having trouble getting lighting to work in program that rotates a cube. All I see is the solid color of my cube, and no shading of any kind. I've read and re-read what to do, and I can't figure out where I'm going wrong... here are the relevant code segments:
void initLighting(void) {
GLfloat mat_properties[] = {1, 1, 0, 1.0};
GLfloat white_light[] = {1.0, 1.0, 1.0, 1.0};
GLfloat light_position[] = {5.0, 5.0, -5.0, 1.0 };

glLightfv(GL_LIGHT0, GL_POSITION, light_position);
glLightfv(GL_LIGHT0, GL_DIFFUSE, white_light);
glLightfv(GL_LIGHT0, GL_AMBIENT, white_light);

/* object color determined here */
glMaterialfv(GL_FRONT, GL_AMBIENT, mat_properties);
glMaterialfv(GL_FRONT, GL_DIFFUSE, mat_properties);

glEnable(GL_LIGHTING);
glEnable(GL_LIGHT0);
}

and in the display callback func:
glBegin(GL_QUADS);
for(int i = 0; i < 6; i++)
{
GLfloat d1[3], d2[3], norm[3];
for(int j = 0; j < 3; j++)
{
d1[j] = vdata[qindices[0]-1][j] - vdata[qindices[i][1]-1][j];
d2[j] = vdata[qindices[i][2]-1][j] - vdata[qindices[i][1]-1][j];
}
normcrossprod(d1,d2,norm);
glNormal3fv(norm);
glVertex3fv(&vdata[qindices[i][0]-1][0]);
glVertex3fv(&vdata[qindices[i][1]-1][0]);
glVertex3fv(&vdata[qindices[i][2]-1][0]);
glVertex3fv(&vdata[qindices[i][3]-1][0]);
}

glEnd();

Thanks edit: there should be an [ i ] after the first qindices, but I don't know how to turn off the codes Edited by - Chupa on March 22, 2002 6:04:47 PM

##### Share on other sites
Well, instead of setting the light to white 1.0,1.0,1.0, try setting it to something like 0.01,0.01,0.01 or so... That oughta do the trick...

Although I''m not sure what the standard of these should be: on one computer I have to set the values to 100 times greater than the original value. So on the one computer I use 10 (x3) and on the other I use 0.01 (x3) with the same graphical result.

Well, I kinda drifted from your problem to my problem but just try setting the amplitude of the color to something way less.

Also try setting the ambient to zero, this way only the light from the source will be applied, and when the source doesn''t reach a face, it must be black (if no other lights are applied)

/G

##### Share on other sites
Its your ambient light. Ambient lighting is basically how much default light you get, whether you are lit by a light or not. Since you set this to 100%, everything will be 100% lit whether its facing the light or not. Try .2, .2, .2 gray instead.

##### Share on other sites
how can i do an enviroment light like inthose quake engines?

##### Share on other sites
your normal is not normalized to length 1.
could be a problem
either normalize it, or glEnable(GL_NORMALIZE) to let gl do it automatically..

##### Share on other sites
Well, besiror. I''m not sure what kind of technic they use in quake to get shadows and nice looking lighting effects. I think it''s lightmap or so, I haven''t seen quake in a long time. If you mean quake 1 I''m almost sure they use only ordinary vertex map...

If you really want state of the art lighting effects, you should check out developer.nvidia.com

/G

##### Share on other sites
My problem was what Gee said, the color was too bright.

Thanks!

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text:
### Information

 Deadline No deadline Submission limit No limitation

## [Part4] Element Constraint

##### Question 1: 1D Element: Domain Consistency

Assume an Element constraint for $$T[y]=z$$, with:

• $$D(y) = \{0,1,2,3,4\}$$
• $$T = [5,5,5,10,8]$$
• $$D(z) = \{5,10\}$$

What is the domain of $$y$$ after a domain-consistent filtering?

##### Question 2: 2D Element: Domain Consistency

Assume an Element constraint for $$T[x][y]=z$$, with:

• $$D(x) = \{0,1,2\}$$
• $$D(y) = \{3\}$$
• $$T = [[1,8,9,6],[1,9,2,4],[9,8,9,8],[1,9,2,5]]$$
• $$D(z) = \{2,...,6\}$$

What is the domain of $$z$$ after a domain-consistent filtering?

##### Question 3: 1D Element: Domain Consistency

Assume an Element constraint for $$T[y]=z$$, with:

• $$D(y) = \{0,1,2,3,4\}$$
• $$T = [5,5,5,10,8]$$
• $$D(z) = \{5,6,7,8,9,10\}$$

What is the domain of $$z$$ after a domain-consistent filtering?

##### Question 4: Stable Matching: Modeling

In the stable matching problem, the constraint "the company of the student of company c is c" is modeled with:


text:
# ! Undefined control sequences

I am sorry that my question title is not clear. I am a beginner using of latex. I copied this template from internet. After modification, I receive this error, which I can't figure out.

! Undefined control sequence.
\f@ncyoch ...hmwkClassInstructor \ \hmwkClassTime
): \hmwkTitle \strut
l.211 \]

This is my text:

\documentclass{article}

\usepackage{fancyhdr}
\usepackage{extramarks}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{tikz}
\usepackage[plain]{algorithm}
\usepackage{algpseudocode}

\usetikzlibrary{automata,positioning}

\topmargin=-0.45in
\evensidemargin=0in
\oddsidemargin=0in
\textwidth=6.5in
\textheight=9.0in

\pagestyle{fancy}
\lfoot{\lastxmark}
\cfoot{\thepage}

\renewcommand\footrulewidth{0.4pt}

\setlength\parindent{0pt}

\nobreak\extramarks{}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
}

\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
\stepcounter{#1}
\nobreak\extramarks{Problem \arabic{#1}}{}\nobreak{}
}

\setcounter{secnumdepth}{0}
\newcounter{partCounter}
\newcounter{homeworkProblemCounter}
\setcounter{homeworkProblemCounter}{1}
\nobreak\extramarks{Problem \arabic{homeworkProblemCounter}}{}\nobreak{}

\newenvironment{homeworkProblem}[1][-1]{
\ifnum#1>0
\setcounter{homeworkProblemCounter}{#1}
\fi
\section{Problem \arabic{homeworkProblemCounter}}
\setcounter{partCounter}{1}
}{
}

\newcommand{\hmwkTitle}{Homework\ \#4}
\newcommand{\hmwkDueDate}{May 07, 2015}
\newcommand{\hmwkClass}{Laser Physics I}
\newcommand{\hmwkClassInstructor}{Professor Kyuman Cho}
\newcommand{\hmwkAuthorName}{Seng Sopheak}

\title{
\vspace{2in}
\textmd{\textbf{\hmwkClass:\ \hmwkTitle}}\\
\normalsize\vspace{0.1in}\small{Due\ on\ \hmwkDueDate\ at 23:59}\\
\vspace{0.1in}\large{\textit{\hmwkClassInstructor}}
\vspace{3in}
}

\author{\textbf{\hmwkAuthorName}}
\date{}

\renewcommand{\part}[1]{\textbf{\large Part \Alph{partCounter}}\stepcounter{partCounter}\\}

\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}

% For derivatives
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}

% For partial derivatives
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}

% Integral dx
\newcommand{\dx}{\mathrm{d}x}

% Alias for the Solution section header
\newcommand{\solution}{\textbf{\large Solution}}

% Probability commands: Expectation, Variance, Covariance, Bias
\newcommand{\E}{\mathrm{E}}
\newcommand{\Var}{\mathrm{Var}}
\newcommand{\Cov}{\mathrm{Cov}}
\newcommand{\Bias}{\mathrm{Bias}}

\begin{document}

\maketitle

\pagebreak

\begin{homeworkProblem}
\textbf{A Plano-Concave Resonator.} When mirror $1$ is planar $\left(R_1=\infty\right)$, determine the confinement condition and the depth of focus, as well as the beam width at the waist and at each of the mirrors, as a function of $\frac{d}{\left|R_2\right|}$.\\

\textbf{Solution}

\textbf{Confinement Condition}
\\
Note that:

$\begin{split} R_1&=z_1\left[z_1+\left(\frac{z_0}{z_1}\right)^2\right] \\ &=\infty \\ \Rightarrow z_1&=0 \\ z_2&=z_1+d \\&=0+d \\ &=d\\ \end{split}$
Confinement Condition:\\
$\begin{array}{r c l}\quad 0\leq&\left(1+\frac{d}{R_1}\right)\left(1+\frac{d}{R_2}\right)&\leq1\\ 0\leq&\left(1+\frac{d}{\infty}\right)\left(1+\frac{d}{\left|R_2\right|}\right)&\leq1\\ 0\leq&\left(1+\frac{d}{\left|R_2\right|}\right)&\leq1\\ \Rightarrow1\geq&\frac{d}{\left|R_2\right|}&\geq0 \end{array}$
\textbf{Depth of Focus}
\\
From\\
$\begin{split} |R_2|&=z_2\left[z_2+\left(\frac{z_0}{z_2}\right)^2\right] \\ \Rightarrow z_0^2&=z_2R_2-z_2^2 \\ &=dR_2-d^2 \\ \Rightarrow z_0&=\sqrt{dR_2-d^2} \\ \textrm{Or }2z_0&=2d\sqrt{\frac{\left|R_2\right|}{d}-1}\\ \end{split}$
\textbf{Beam Width}
\\
At the plane mirror\\
$\begin{split} W\left(z_1\right)&=W_0\sqrt{1+\left(\frac{z_1}{z_0}\right)^2} \\ &=W_0\sqrt{1+\left(\frac{\infty}{z_0}\right)^2} \\ &=W_0 \end{split}$\\
At the concave mirror\\
$\begin{split} W\left(z_2\right)&=W_0\sqrt{1+\left(\frac{z_2}{z_0}\right)^2} \\ &=W_0\sqrt{1+\left(\frac{d}{d\sqrt{\frac{R_2}{d}-1}}\right)^2} \\ &=W_0\sqrt{1+\frac{d}{R_2-d}} \end{split}$

\end{homeworkProblem}

\end{document}

I am sorry to copy too long. I don't know where the mistake if or what it is for that matter. Thank

The error message is pretty clear, \hmwkClassTime is not defined. Therefore define it or remove it.

• Thank! I couldn't have found it without you. Sorry I am a really beginner.
– TBBT
May 7 '15 at 9:30

The original template is called Structured General Purpose in the assignments section of the template site.

We can find

\newcommand{\hmwkClassTime}{10:30am} % Class/lecture time

Since this is used multiple times, empty the argument to prevent the error.

\newcommand{\hmwkClassTime}{} % Class/lecture time

It should be noted, that templates seem to be a good idea to start LaTeX, but they aren't. The LaTeX coding is mostly horrible and a very bad example and should not be used as a starting point.


text:
# Mathematical Models in Epidemiology

###### Fred Brauer, Carlos Castillo-Chavez, Zhilan Feng
Publisher:
Springer
Publication Date:
2019
Number of Pages:
619
Format:
Hardcover
Series:
Texts in Applied Mathematics
Price:
84.99
ISBN:
978-1-4939-9826-5
Category:
Textbook
[Reviewed by
Glenn Ledder
, on
06/14/2020
]
A 2019 book on epidemiology from three of the world’s top experts on the subject could not be more timely.  To the best of my knowledge, none of these experts have been contacted by public health authorities for advice.  Given that, it is worthwhile to start by quoting the first few sentences of the final chapter, titled “Challenges, Opportunities, and Theoretical Epidemiology.”

Lessons learned from the HIV pandemic, SARS in 2003, the 2009 H1N1 influenza pandemic, the 2014 Ebola outbreak in West Africa, and the ongoing Zika outbreaks in the Americas can be framed under a public health policy model that responds after the fact.  Responses often come through reallocation of resources from one disease control effort to a new pressing need.  The operating models of preparedness and response are ill-equipped to prevent or ameliorate disease emergence at global scales.

This 16-chapter book is an 8-fold expansion of the material on epidemiology that appears in the book Mathematical Models in Population Biology and Epidemiology by the first two authors.  It is a detailed treatment of deterministic epidemiology models.  The core of the book consists of a chapter that presents the basic SIS and SIR models, with and without demographics (births and deaths), followed by specific chapters on endemic models (with demographics), epidemic models (without demographics), models with heterogeneous mixing, and models for vector-transmitted diseases.  This core is followed by six chapters that present models for specific diseases of current interest, such as tuberculosis, HIV/AIDS, and Zika.

It would be impossible to review the book in its entirety, nor would a broad superficial review give the reader as much of a sense of the book as a more focused look at a small number of topics.

The standard Kermack-McKendrick epidemic SIR model is presented in Section 2.4.  This model combines a transmission process that moves individuals from susceptible to infected with a removal process that moves individuals from infected to removed.  The assumptions used to quantify these processes in the model are clearly wrong.  Using βSI as the transmission rate assumes that people in a community are like molecules in a gas, moving around randomly and having a normal distribution of contacts, while real communities have contacts based on interpersonal networks with an asymmetrical distribution of contact rates among individuals.  Using αI as the removal rate assumes that the time spent in the infected class is exponentially distributed, which means that more individuals are infected for exactly one day than for exactly five days, even if the mean infectious period is five days.  We often use this standard model without consideration of the amount of error caused by these flawed assumptions, but this book addresses them in some detail.  Chapter 4 begins with a stochastic model that assumes contacts between individuals occur in a network of connections rather than through random mixing.  The development and analysis of the model ultimately leads to discovery of the specific assumptions needed for the model to reduce to the usual mass action transmission rate.  This development clarifies the conditions needed for the mass action assumption to be appropriate.  Later in the chapter, the authors replace the exponential distribution of infection duration with a general distribution and show that the relationship between the fraction of individuals who are never infected and the basic reproductive number is the same for any distribution.  The exponential distribution may yield systematic error in the early stages, but it yields the same ultimate outcome.

Among the models presented for specific diseases are two models for dengue fever.  The first of these modifies the basic model from the core chapter on vector-transmitted diseases by incorporating vertical transmission of the pathogen in the mosquito population.  The second model incorporates an additional class of asymptomatic infectives that has a reduced transmission rate and an increased recovery rate.  The addition of asymptomatics to the dengue model is timely, as a similar change is needed to adapt the standard SEIR model to a model for COVID-19.  The study of both dengue fever models focuses on calculating the basic reproductive number using the direct method of working with the average numbers of transmissions from one human, from one human-infected mosquito, and from one mosquito infected at birth.  The computation of the basic reproduction number using the next generation matrix method is left for the exercises.

No book can be both a gentle introduction for beginners and a serious reference book for specialists; this one is the latter and not the former.  It can be used as a textbook for courses, but anyone who considers it for that purpose should make sure it is a good fit for their students.  I would not hesitate to use it for a graduate-level course, but I would be cautious about using it with undergraduates.  The authors assume a significant level of general mathematical modeling background, which would make the book a difficult read for a typical advanced undergraduate student whose modeling experience is limited to what they have seen in mathematics courses.  A model with multiple parameters, whose behavior is to be studied without specific parameter values, is a quantum leap beyond the typical “application” that contains only one parameter whose value can be fixed by one piece of data.  Sections 2.2 through 2.4, on the SIS model and on SIR models with and without demographics, focus on analytical results (both quantitative and qualitative), while presenting only one figure that shows plots of class populations versus time and only one phase portrait.  On the plus side, the book is well endowed with exercises of varying levels of difficulty, along with a number of sections that outline possible student projects.

As with all first edition books, one should be wary of proofreading errors.  One particularly unfortunate error is at the end of Section 2.3, where the authors derive an equation for the total population N in an SIR model with constant birth rate and disease-related death, $N'=\Lambda - \mu N - \alpha I$, and then discuss the effect of the pathogenicity parameter on the equilibrium population.  Unfortunately, this equation is derived from one a few pages back that uses $d$ as the pathogenicity parameter and $\alpha$ as the mean recovery rate of infectives.  Presumably, this will be corrected in a subsequent printing.

Whether one chooses this book for use in a class or not, every practicing mathematical epidemiologist will want this book to occupy a prominent place in their library.

Glenn Ledder has done research work in combustion theory, groundwater flow, population dynamics, plant life history theory, and plant physiology.  In addition to his research, he has worked in mathematical pedagogy, editing an MAA notes volume and writing textbooks in differential equations and mathematics for the life sciences.


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# What is the relation between min-vertex cover and min-cut in a bipartite graph?

In a bipartite graph, the size of a maximum matching equals the size of the minimum vertex cover.

I want to prove above theorem using max-flow-min-cut theorem. So I am trying to find the relationship between min-cut and min vertex cover, do you think |min-cut| = |min-vertex-cover|? If so, any proof for it?

• The max-flow min-cut theorem applies to a flow network with a source, a sink, and edge weights. A bipartite graph does not have those, by default. So as it stands, your equality does not even make sense. – Misha Lavrov Oct 3 '17 at 5:21
• what if I constructed new graph G` by adding a source and sink vertex to bipartite graph, such that flow from source to any vertex is zero, and any vertex to sink is zero. and all capacity of the new construction is 1. – ajayramesh Oct 3 '17 at 14:31

I think your confusion lies in the correspondence between flows in a bipartite graph and maximum matchings. To find a maximum matching, consider the bipartite graph $$G=(L,R,E)$$ where $$L$$ is the set of vertices on the left side of the bipartition and $$R$$ is the set on the right side. Now, add a new vertex $$s$$ and a directed edge $$(s,\ell)$$ for each $$\ell \in L$$. Also, add a new vertex $$t$$ and a directed edge $$(r,t)$$ for each $$r\in R$$. Orient all of the edges in $$E$$ towards $$T$$ and set every edge in your new graph to have capacity 1. Below is an example of the resulting graph:

Solve for max flow on this graph with $$s$$ as your source and $$t$$ as your terminal. The saturated edges (which were originally in $$E$$) in a max flow for this graph correspond to the edges in a maximum matching, and the flow value is the size of said matching (convince yourself! i.e. assume you have a max matching that is bigger/ smaller than the flow value and see what that implies).

Now, the flow value also corresponds to the size of a minimum vertex cover, but NOT NECESSARILY the size of a minimum cut. This is because we added a source and a sink node, and the edges from $$s$$ to $$L$$ and from $$R$$ to $$t$$ which could now be part of our min cut in this new graph. Hope this clears up some confusion.

If I'm interpreting your question and comment correctly, you mean a minimum set of edges such that their deletion would make the graph disconnected. Consider the path with $4$ vertices, which is bipartite. You'll need at least two vertices to hit every edge, but deleting any edge will make the graph disconnected, so $|\mathrm{min\;cut}|\neq|\mathrm{min\;vertex\;cover}|$.

Contrary to the suggestions of the other users, I believe this can be proved (the counterexample provided does not address the OP's problem as OP clarified it in the comments).

Let the original graph be $$G$$, the new graph be $$G^\prime$$, the vertices on the left be $$V_1$$, the vertices on the right be $$V_2$$, and the source and sink be $$s$$ and $$t$$ respectively.

$$|\text{min cut}| \leq |\text{min vertex cover}|$$:

Let $$C_1 \cup C_2$$, where $$C_1 \subseteq V_1, C_2 \subseteq V_2$$, be the smallest vertex cover. Consider the following cut:

$$S = \{s\} \cup (V_1 \setminus C_1 ) \cup C_2$$

Then

$$\bar{S} = \{t\} \cup C_1 \cup (V_2 \setminus C_2)$$

Note that none of the arcs from $$V_1$$ to $$V_2$$ are present in the cut. We can see this as follows. If an arc starts from a vertex in $$V_1$$ inside the vertex cover, then it is not in $$S$$ in the first place. If an arc starts from a vertex in $$V_1$$ not in the vertex cover, then it must end in a vertex in $$V_2$$ in the cover (otherwise we would not have a vertex cover). Therefore, the arc must not be in the cut, since it connects two vertices within $$S$$.

As a result, the only arcs in the cut are those from $$\{s\}$$ to $$C_1$$ and from $$C_2$$ to $$\{t\}$$. The size of this cut is $$|C_1 \cup C_2|$$, or the size of the minimum vertex cover. Therefore, the size of the min cut is at most the size of the min vertex cover.

$$|\text{min cut}| \geq |\text{min vertex cover}|$$:

Let the minimum cut be $$(S,\bar{S})$$. We express the cut as follows:

$$S = \{s\} \cup (V_1 \setminus C_1 ) \cup C_2$$

$$\bar{S} = \{t\} \cup C_1 \cup (V_2 \setminus C_2)$$

where $$C_1 \subseteq V_1$$ and $$C_2 \subseteq V_2$$. Clearly, the arcs inside this cut consist of all the arcs from $$\{s\}$$ to $$C_1$$, from $$V_2 \cap C_2$$ to $$\{t\}$$, and from $$V_1 \setminus C_1$$ to $$V_2 \setminus C_2$$.

Consider the set $$C_1 \cup C_2$$. We may expand this set to a vertex cover by adding all the vertices of $$V_1 \setminus C_1$$ for which an arc from it to $$V_2 \setminus C_2$$ exists. Obviously the size of this set is the same as the size of the min cut. So the size of the minimum vertex cover is at most the size of the minimum cut and we are done.


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# zbMATH — the first resource for mathematics

Minimal vertex degree sum of a 3-path in plane maps. (English) Zbl 0906.05017
A plane map is normal if every face and every vertex of the map is incident with at least three edges. Let $$w_k$$ be the minimum degree sum, in a graph $$G$$, of the vertices on a path of order $$k$$ in $$G$$. The author shows that, for normal plane maps: (1) If $$w_2= 6$$, then $$w_3$$ can be arbitrarily large. (2) If $$w_2> 6$$, then either $$w_3\leq 18$$ or there is a vertex of degree at most 15 adjacent to two vertices of degree 3. (3) If $$w_2>7$$, then $$w_3\leq 17$$.

##### MSC:
 05C10 Planar graphs; geometric and topological aspects of graph theory 05C38 Paths and cycles
##### Keywords:
plane map; minimum degree sum; path
Full Text:


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Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 5.2 - Problem 8
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# ?In Exercises 7–12, use sigma notation to write the sum. $$\frac{9}{1+1}+\frac{9}{1+2}+\frac{9}{1+3}+\cdots+\frac{9}{1+14}$$

ISBN: 9781285774770 141

## Solution for problem 8 Chapter 5.2

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Problem 8

In Exercises 7–12, use sigma notation to write the sum.

$$\frac{9}{1+1}+\frac{9}{1+2}+\frac{9}{1+3}+\cdots+\frac{9}{1+14}$$

Text Transcription:

\frac{9}{1+1}+\frac{9}{1+2}+\frac{9}{1+3}+\cdots+\frac{9}{1+14}

Step-by-Step Solution:

Step 1 of 5) Figure 1.36 The radian measure of the central angle ACB is the number u = s>r. For a unit circle of radius r = 1, u is the length of arc AB that central angle ACB cuts from the unit circle.

Step 2 of 2

##### ISBN: 9781285774770

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# Thompson sampling

"Bayesian control rule" redirects here. For other uses, see Bayesian control rule (disambiguation).

In artificial intelligence, Thompson sampling,[1] named after William R. Thompson, is a heuristic for choosing actions that addresses the exploration-exploitation dilemma in the multi-armed bandit problem. It consists in choosing the action that maximizes the expected reward with respect to a randomly drawn belief.

## Description

Consider a set of contexts $\mathcal{X}$, a set of actions $\mathcal{A}$, and rewards in $\mathbb{R}$. In each round, the player obtains a context $x \in \mathcal{X}$, plays an action $a \in \mathcal{A}$ and receives a reward $r \in \mathbb{R}$ following a distribution that depends on the context and the issued action. The aim of the player is to play actions such as to maximize the cumulative rewards.

The elements of Thompson sampling are as follows:

1. a set $\Theta$ of parameters $\theta$;
2. a prior distribution $P(\theta)$ on these parameters;
3. past observations triplets $\mathcal{D} = \{(x; a; r)\}$;
4. a likelihood function $P(r|\theta,a,x)$;
5. a posterior distribution $P(\theta|\mathcal{D}) \propto P(\mathcal{D}|\theta)P(\theta)$, where $P(\mathcal{D}|\theta)$ is the likelihood function.

Thompson sampling consists in playing the action $a^\ast \in \mathcal{A}$ according to the probability that it maximizes the expected reward, i.e.

$\int \mathbb{I}[\mathbb{E}(r|a,x,\theta) = \max_{a'} \mathbb{E}(r|a',x,\theta)] P(\theta|\mathcal{D}) \, d\theta,$

where $\mathbb{I}$ is the indicator function.

In practice, the rule is implemented by sampling, in each round, a parameter $\theta^\ast$ from the posterior $P(\theta|\mathcal{D})$, and choosing the action $a^\ast$ that maximizes $\mathbb{E}[r|\theta^\ast,a^\ast,x]$, i.e. the expected reward given the parameter, the action and the current context. Conceptually, this means that the player instantiates his beliefs randomly in each round, and then he acts optimally according to them.

## History

Thompson sampling was originally described in an article by Thompson from 1933 [1] but has been largely ignored by the artificial intelligence community. It was subsequently rediscovered numerous times independently in the context of reinforcement learning.[2][3][4][5][6][7] A first proof of convergence for the bandit case has been shown in 1997.[2] The first application to Markov decision processes was in 2000.[4] A related approach (see Bayesian control rule) was published in 2010.[3] In 2010 it was also shown that Thompson sampling is instantaneously self-correcting.[7] Asymptotic convergence results for contextual bandits were published in 2011.[5] Thompson sampling has also been applied to A/B testing in website design and online advertising.[8] Recently, Thompson sampling has formed the basis for accelerated learning in decentralized decision making.[9]

## Relationship to other approaches

### Probability matching

Probability matching is a decision strategy in which predictions of class membership are proportional to the class base rates. Thus, if in the training set positive examples are observed 60% of the time, and negative examples are observed 40% of the time, the observer using a probability-matching strategy will predict (for unlabeled examples) a class label of "positive" on 60% of instances, and a class label of "negative" on 40% of instances.

### Bayesian control rule

A generalization of Thompson sampling to arbitrary dynamical environments and causal structures, known as Bayesian control rule, has been shown to be the optimal solution to the adaptive coding problem with actions and observations.[3] In this formulation, an agent is conceptualized as a mixture over a set of behaviours. As the agent interacts with its environment, it learns the causal properties and adopts the behaviour that minimizes the relative entropy to the behaviour with the best prediction of the environment's behaviour. If these behaviours have been chosen according to the maximum expected utility principle, then the asymptotic behaviour of the Bayesian control rule matches the asymptotic behaviour of the perfectly rational agent.

The setup is as follows. Let $a_1, a_2, \ldots, a_T$ be the actions issued by an agent up to time $T$, and let $o_1, o_2, \ldots, o_T$ be the observations gathered by the agent up to time $T$. Then, the agent issues the action $a_{T+1}$ with probability:[3]

$P(a_{T+1}|\hat{a}_{1:T}, o_{1:T}),$

where the "hat"-notation $\hat{a}_t$ denotes the fact that $a_t$ is a causal intervention (see Causality), and not an ordinary observation. If the agent holds beliefs $\theta \in \Theta$ over its behaviors, then the Bayesian control rule becomes

$P(a_{T+1}|\hat{a}_{1:T}, o_{1:T}) = \int_{\Theta} P(a_{T+1}|\theta, \hat{a}_{1:T}, o_{1:T}) P(\theta|\hat{a}_{1:T}, o_{1:T}) \, d\theta$,

where $P(\theta|\hat{a}_{1:T}, o_{1:T})$ is the posterior distribution over the parameter $\theta$ given actions $a_{1:T}$ and observations $o_{1:T}$.

In practice, the Bayesian control amounts to sampling, in each time step, a parameter $\theta^\ast$ from the posterior distribution $P(\theta|\hat{a}_{1:T}, o_{1:T})$, where the posterior distribution is computed using Bayes' rule by only considering the (causal) likelihoods of the observations $o_1, o_2, \ldots, o_T$ and ignoring the (causal) likelihoods of the actions $a_1, a_2, \ldots, a_T$, and then by sampling the action $a^\ast_{T+1}$ from the action distribution $P(a_{T+1}|\theta^\ast,\hat{a}_{1:T},o_{1:T})$.

## References

1. ^ a b Thompson, William R. "On the likelihood that one unknown probability exceeds another in view of the evidence of two samples". Biometrika, 25(3–4):285–294, 1933.
2. ^ a b J. Wyatt. Exploration and Inference in Learning from Reinforcement. Ph.D. thesis, Department of Artificial Intelligence, University of Edinburgh. March 1997.
3. ^ a b c d P. A. Ortega and D. A. Braun. "A Minimum Relative Entropy Principle for Learning and Acting", Journal of Artificial Intelligence Research, 38, pages 475–511, 2010.
4. ^ a b M. J. A. Strens. "A Bayesian Framework for Reinforcement Learning", Proceeedings of the Seventeenth International Conference on Machine Learning, Stanford University, California, June 29–July 2, 2000, http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.140.1701
5. ^ a b B. C. May, B. C., N. Korda, A. Lee, and D. S. Leslie. "Optimistic Bayesian sampling in contextual-bandit problems". Technical report, Statistics Group, Department of Mathematics, University of Bristol, 2011.
6. ^ Chapelle O. and Li, L. "An Empirical Evaluation of Thompson Sampling". NIPS, 2011.[full citation needed]
7. ^ a b O.-C. Granmo. "Solving Two-Armed Bernoulli Bandit Problems Using a Bayesian Learning Automaton", International Journal of Intelligent Computing and Cybernetics, 3 (2), 2010, 207-234.
8. ^ Ian Clarke. "Proportionate A/B testing", September 22nd, 2011, http://blog.locut.us/2011/09/22/proportionate-ab-testing/
9. ^ Granmo, O. C.; Glimsdal, S. (2012). "Accelerated Bayesian learning for decentralized two-armed bandit based decision making with applications to the Goore Game". Applied Intelligence. doi:10.1007/s10489-012-0346-z.


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# I Thermodynamics: Gas Expansion with Piston Friction

#### Chestermiller

Mentor
I'm thinking of the following comparison:

System 1: massless piston with friction force F, and a pile of small masses totaling M sitting on top of the piston initially

System 2: frictionless piston having mass m, such that mg for this system is equal to F for system 1, and the same pile of small masses totaling M sitting on top of the piston initially

The initial states of these two systems will be the same, but System 2 will experience an adiabatic reversible expansion and System 1 will experience an adiabatic irreversible expansion when the same pile of small masses is removed quasi-statically from the top of each piston.

#### Robert Davidson

I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Yes, I would very much appreciate it if you could provide all the analysis.

For the process to be quasi-static, we cannot let the piston rise on its own…..

I was not thinking of a quasi-static process. What I had in mind was to compare the final volume for two irreversible constant pressure adiabatic expansions, one with sliding friction and one without.

#### Chestermiller

Mentor
I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Yes, I would very much appreciate it if you could provide all the analysis.
I'll post this in a little while.

For the process to be quasi-static, we cannot let the piston rise on its own…..

I was not thinking of a quasi-static process. What I had in mind was to compare the final volume for two irreversible constant pressure adiabatic expansions, one with sliding friction and one without.
So you would consider the same comparison as in my previous post, except with sudden removal of mass M at time zero?

#### Chestermiller

Mentor
The starting equations for the frictional analysis I alluded to in post #20 is:
$$dU=nC_vdT=\frac{F}{A}dV-\frac{nRT}{V}dV\tag{1}$$and $$dU=nC_vdT=-\frac{f}{A}dV\tag{2}$$Dividing Eqn. 1 by $nC_vT$ yields:
$$\frac{dT}{T}=\frac{FdV}{AnC_vT}-\frac{R}{C_v}\frac{dV}{V}=\left[\frac{FV}{AnC_vT}-\frac{R}{C_v}\right]\frac{dV}{V}\tag{3}$$
In Eqn. 3, $$\frac{FV}{AnC_vT}=\frac{FV_0}{nC_vAT_0}\frac{(V/V_0)}{(T/T_0)}=\frac{FR}{AC_v}\frac{V_0}{nRT_0}\frac{(V/V_0)}{(T/T_0)}$$
But, from the ideal gas law, $$\frac{nRT_0}{V_0}=P_0$$
Therefore, $$\frac{FV}{AnC_vT}=\frac{F}{P_0A}\frac{C_v}{R}\frac{(V/V_0)}{(T/T_0)}\tag{4}$$
Substituting Eqn. 4 into Eqn. 3 then yields:
$$\frac{dT}{T}=\frac{C_v}{R}\left[\frac{F}{P_0A}\frac{(V/V_0)}{(T/T_0)}-1\right]\frac{dV}{V}=(\gamma -1)\left[\frac{F}{P_0A}\frac{(V/V_0)}{(T/T_0)}-1\right]\frac{dV}{V}\tag{5}$$
We now make the following dimensionless substitutions:
$$\bar{T}=T/T_0$$
$$\bar{V}=V/V_0$$
This then yields:
$$\frac{d\ln{\bar{T}}}{d\ln{\bar{V}}}=-(\gamma -1)\left[1-\frac{F}{P_0A}e^{(ln{\bar{V}}-\ln{\bar{T}})}\right]\tag{6}$$
This is the differential equation I integrated to get the results in post #20.
To get the work terms, I did the following: From Eqn. 2, if I divide by $nC_vT_0$ and integrate, I obtain:
$$\frac{\left[\int_{V_0}^V{\frac{f}{A}dV}\right]}{nC_vT_0}=1-\bar{T}$$
Similarly, for the integral of the gas pressure, I get:
$$\frac{\left[\int_{V_0}^V{PdV}\right]}{nC_vT_0}=(\gamma-1)\frac{F}{P_0A}(\bar{V}-1)+(1-\bar{T})\tag{7}$$

#### Robert Davidson

Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights plus atmospheric pressure.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob

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#### Chestermiller

Mentor
Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob
Bob,

Although I don't think that the two cases are really going to be comparable, I think each of them is interesting on its own. So please proceed and let's see what you come up with. (Please note that, in both these cases, the gas experiences a non-quasistatic deformation, and entropy is generated within the gas in each case).

Chet

#### Chestermiller

Mentor
I've been able to analytically solve the differential equation (Eqn. 6) of post #29 for the dimensionless temperature of the gas as a function of the dimensionless volume ratio. The result, which agrees with the numerical solution I presented in post #20 is:
$$\bar{T}=\frac{(1-\xi)}{\bar{V}^{(\gamma-1)}}+\xi \bar{V}$$where $$\xi=\frac{(\gamma-1)}{\gamma}\frac{F}{P_0A}$$

#### Robert Davidson

Hi Chet,

Haven’t forgot this. So far only tackled the easy case (system 2). Here’s what I have so far.

Going to assume dealing with ideal gas (air) with the following initial conditions and properties:

$P_{i}=10atm$, $T_{i}=300K$, $n=1$, $C_{v}=0.2053\frac{l.atm}{mol.K}$, $R_{u}=0.08205\frac{l.atm}{mol.K}$, $k=1.4$

Applying the ideal gas equation to get the initial volume:

$$V_{i}=\frac{nRT_{i}}{P_{i}}=\frac{(1)(0.08205)(300)}{10}=2.46 l$$

Weight is abruptly removed to reduce the external pressure by one half, or to 5 atm, and the gas allowed to expand non-quasi-statically at constant pressure until equilibrium is reached.

Since the final pressure is known, we can (1) determine the relationship between the final temperature and volume by applying the general gas equation to the initial and final states and (2) substitute the relationship into the first law equation for a constant pressure adiabatic process.

Applying the general gas equation:

$$\frac{P_{i}V_{i}}{T_{i}}= \frac{P_{f}V_{f}}{T_{f}}$$

$$T_{f}=\frac{P_{f}V_{f}T_{i}}{P_{i}V_{i}}$$

$$T_{f}=\frac{(5)(V_{f})(300)}{(10)(2.46)}=61V_{f}$$

Applying the first law for an adiabatic process:

$$\Delta U=-W$$

$$nC_{v}(T_{f}-T_{i})=-P_{f}(V_{f}-V_{i})$$

Substituting $T_{f}=61V_{f}$ and the other known values

$$(1)(0.2053)[61V_{f}-300]=-5(V_{f}-2.46)$$

Which gives us $V_{f}=4.22l$ and $T_{f}=257K$, and the work done is

$$W=5(4.22-2.46)l.atm=892j$$

Comparing to the final volume and work done for a reversible adiabatic expansion with the same initial conditions and same final pressures:

$$P_{i}V_{i}^{k}= P_{f}V_{f}^{k}$$

$$V_{f}=6l$$

$$W_{rev}=\frac{(P_{f}V_{f}-P_{i}V_{i})}{1-k}$$

$$W_{rev}= 13.5 l.atm=1368j$$

Which, as expected, is greater than the irreversible work.

The next step is to tackle system 1, which adds piston friction. I would like to bring system 2 to the same final pressure and then compare the work on the surroundings to system 2. I recall that in your analysis of the quasi-static expansion with friction it was necessary to apply an external tensile force to overcome static friction in order to reach the same volume as the non-friction system. It appears that will again be the case here, except it would be in order to reach the same final pressure.

I was thinking about a possible alternative approach, such as assuming the surface of the cylinder at some point changes to a smooth service prior to reaching the maximum static friction force so that the expansion could be completed, albeit without friction work done at the end. But that would probably create more complications than it's worth.

#### Chestermiller

Mentor
I approached this problem differently but, of course, arrived at the same results. My focus was on the final temperature, rather than the final volume.
$$nC_v(T_f-T_i)=-P_f(V_f-V_i)=P_f\left(\frac{nRT_f}{P_f}-\frac{nRT_i}{P_i}\right)$$
Solving for $T_f$ then yields:
$$T_f=\left[\frac{1}{\gamma}+\left(1-\frac{1}{\gamma}\right)\frac{P_f}{P_i}\right]T_i$$For $\gamma=1.4$ and $P_f/P_i=0.5$, this yields $T_f=0.8571T_i$. And, for $T_i=300\ K$, $T_f=257.1K$.

The work is given by: $$W=-\Delta U=-nC_v(T_f-T_i)=-(1)(2.5)(8.314)(257.1-300)=892\ J$$
It'll be interesting to compare the approach and results you get for system 2 with those that I obtained.

#### Chestermiller

Mentor
Here's an interesting paradox for you. As $P_f/P_i\rightarrow 0$, $$\frac{T_f}{T_i}\rightarrow\frac{1}{\gamma}$$and then $$W\rightarrow nC_vT_i\left(1-\frac{1}{\gamma}\right)>0$$But how can this be if $P_f/P_i\rightarrow 0$ implies free expansion.

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#### Robert Davidson

Chet
Interesting. Going away on a two week cruise so won't get back to this till I get back. But I'll be thinking about it on the beaches of Barbados!! (Ha Ha).

#### Chestermiller

Mentor
Chet
Interesting. Going away on a two week cruise so won't get back to this till I get back. But I'll be thinking about it on the beaches of Barbados!! (Ha Ha).
Don't laugh. You and I both know that you actually will be thinking about it, even on the beaches of Barbados.

#### Chestermiller

Mentor
To complete this thread, here is the beginning of my solution for system 2 in which there is a friction force F.

Equation for the initial force balance on the piston: $$P_{gi}A-F-P_{atm,i}A=0$$where A is the piston cross sectional area, F is the friction force, $P_{gi}$ is the initial equilibrium gas pressure, and $P_{atm,i}$ is the initial atmospheric pressure.

Equation for the force balance on the piston during the expansion: $$F_g-F-\frac{P_{atm,i}}{2}A=0$$
where $F_g$ is the force that the gas exerts on the inside piston face during this irreversible expansion.

#### Robert Davidson

Hi Chester,

Finally getting back to this.

Thanks for giving the beginning of the solution for system 2. Before I go too far down the road, I need to make sure I’m on the same page with you.

I have no problem with the force balance during expansion (your second equation). I am having difficulties with the force balance for the initial state (the first equation).

I would like to see the initial gas pressure exactly equal to the externally applied pressure (atmosphere plus weights= 10 atm) as we did for system 1. Since there would be no net externally applied force, there would be no fiction force, $F$, to oppose it. In other words, $F=0$ initially.

Once the externally applied force is halved, the difference between the force of the gas and the externally applied force will be the fiction force $F$, as presented in the second equation (with external pressure = 5 atm).

Do you see any issues with this approach?

#### Chestermiller

Mentor
Hi Chester,

Finally getting back to this.

Thanks for giving the beginning of the solution for system 2. Before I go too far down the road, I need to make sure I’m on the same page with you.

I have no problem with the force balance during expansion (your second equation). I am having difficulties with the force balance for the initial state (the first equation).

I would like to see the initial gas pressure exactly equal to the externally applied pressure (atmosphere plus weights= 10 atm) as we did for system 1. Since there would be no net externally applied force, there would be no fiction force, $F$, to oppose it. In other words, $F=0$ initially.

Once the externally applied force is halved, the difference between the force of the gas and the externally applied force will be the fiction force $F$, as presented in the second equation (with external pressure = 5 atm).

Do you see any issues with this approach?
I find this kind of confusing. We are saying that, initially, there is no friction force, but, as soon as we drop the external pressure, friction kicks in at full force during the entire time that the piston is sliding. And it's still present at the very end. I guess this would be OK, under our assumption that the coefficients of static and kinetic friction are equal. But, during the expansion, the friction force would have to be less than 5 atm times the piston area, or the gas would not be able to expand. OK. Let's see where that takes us.

Chet

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"Thermodynamics: Gas Expansion with Piston Friction"

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