原題與已知條件
完整prime field Z_p運算。操作=fielddeep-affine-solve;資料={"p":89,"a":24,"b":24,"c":47}。請依 [a,b,c,inverse,solution,substitution,unique] 欄序回答。
G001 — 完整prime field Z_p運算。操作=fielddeep-affine-solve;資料={"p":89,"a":24,"b":24,"c":47}。請依 [a,b,c,inverse,solution,substitution,unique] 欄序回答。
逐步推導
S001 · implies
操作前
source roles = [qf-mod31-fielddeep-affine-solve, [a, b, c, domain, operation, p]]
操作後
derived = [fielddeep-affine-solve, Exact_prime_field_Zp]
條件:無
原bank的operation是LOCAL hyphen,keys與domain各自精確匹配。欄位順序及數字/Boolean/字串角色依來源,所有有理數文字整數省略/1;不改原題與英文、MKO、AST或狀態。
步驟識別與依賴
規則:R66_EXACT_source_bank_keys_local_operation_domain · 依賴:G001
S002 · implies
操作前
source roles = [89, [2, 3, 4, 5, 6, 7, 8, 9]]
操作後
derived = [[1, 2, 1, 4, 5, 5, 1, 8], 是]
條件:無
完整試除2到floor(sqrt(p))無因數;若p合成必有其中因數。2也合法。只限定本題Z_p,不把一般有限域都說成prime order或把有限環都說成域。
步驟識別與依賴
規則:Field_PRIME_complete_trial_divisor_qualification · 依賴:G001, S001
S003 · calculation
操作前
Field ORIGINAL a times proposed inverse := mul(24, 26)
操作後
Field ORIGINAL a times proposed inverse = 624
條件:無
按本步原操作數與次序精確運算;來源有理數的pre-reduction界另行保留,不用浮點或偷偷跨bank放寬。
步驟識別與依賴
規則:exact_rational_arithmetic · 依賴:G001, S002
S004 · implies
操作前
source roles = [inverse substitution, 624, 89]
操作後
derived = [7, 1]
條件:無
以整數Euclidean division核value=n*quotient+remainder與0≤remainder<n,負代表元也正規化;不是浮點餘數。
步驟識別與依賴
規則:Finite_MOD_euclidean_quotient_remainder · 依賴:G001, S003
S005 · calculation
操作前
Field ORIGINAL c minus b := sub(47, 24)
操作後
Field ORIGINAL c minus b = 23
條件:無
按本步原操作數與次序精確運算;來源有理數的pre-reduction界另行保留,不用浮點或偷偷跨bank放寬。
步驟識別與依賴
規則:exact_rational_arithmetic · 依賴:G001, S004
S006 · calculation
操作前
Field ORIGINAL inverse times c minus b := mul(26, 23)
操作後
Field ORIGINAL inverse times c minus b = 598
條件:無
按本步原操作數與次序精確運算;來源有理數的pre-reduction界另行保留,不用浮點或偷偷跨bank放寬。
步驟識別與依賴
規則:exact_rational_arithmetic · 依賴:G001, S005
S007 · implies
操作前
source roles = [affine solution, 598, 89]
操作後
derived = [6, 64]
條件:無
以整數Euclidean division核value=n*quotient+remainder與0≤remainder<n,負代表元也正規化;不是浮點餘數。
步驟識別與依賴
規則:Finite_MOD_euclidean_quotient_remainder · 依賴:G001, S006
S008 · calculation
操作前
Field ORIGINAL a times solution := mul(24, 64)
操作後
Field ORIGINAL a times solution = 1536
條件:無
按本步原操作數與次序精確運算;來源有理數的pre-reduction界另行保留,不用浮點或偷偷跨bank放寬。
步驟識別與依賴
規則:exact_rational_arithmetic · 依賴:G001, S007
S009 · calculation
操作前
Field ORIGINAL a solution plus b := sum(1536, 24)
操作後
Field ORIGINAL a solution plus b = 1560
條件:無
按本步原操作數與次序精確運算;來源有理數的pre-reduction界另行保留,不用浮點或偷偷跨bank放寬。
步驟識別與依賴
規則:exact_rational_arithmetic · 依賴:G001, S008
S010 · implies
操作前
source roles = [equation substitution, 1560, 89]
操作後
derived = [17, 47]
條件:無
以整數Euclidean division核value=n*quotient+remainder與0≤remainder<n,負代表元也正規化;不是浮點餘數。
步驟識別與依賴
規則:Finite_MOD_euclidean_quotient_remainder · 依賴:G001, S009
S011 · implies
操作前
source roles = [89, 24, 24, 47, 26]
操作後
derived = [64, 47, 是]
條件:無
p為prime且a非0故可逆;先減b再乘inverse,代回同一模p等式。若兩解x,z,a(x−z)=0,乘inverse得x=z,故唯一;不除以0。
步驟識別與依賴
規則:Field_INVERT_nonzero_coefficient_and_UNIQUENESS · 依賴:G001, S010