PROBLEM TO BE SOLVED: To detect falsification during secret calculation using a plurality of secrecy distributions.;SOLUTION: A secrecy calculation device 1n inputs distribution values [a0], ..., [aM-1]. A random number generation section 12 calculates distribution values [r0], ..., [rJ-1]. A randomization section 13 calculates a distribution value [amrj] obtained by integrating a distribution value [am] and a distribution value [rj] to generate a randomized distribution value am≒[am],[amrj]. A secrecy calculation section 14 calculates a function value [F([a0], ..., [aM-1])] while including a calculation object and a randomized distribution value as a calculation result in a checksum Cj. A synchronization section 15 stands by until the secrecy calculation using all the secrecy distributions is terminated. A correctness proving section 16 validates whether or not a distribution value [φj] obtained by multiplying a sum of the distribution values [φj][f0], ..., [fμj-1] included in the checksum Cj by the distribution value [rj] is equal to a distribution value [ψj] as a sum of the distribution values [f0rj], ..., [fμj-1rj] included in the checksum Cj.;SELECTED DRAWING: Figure 2;COPYRIGHT: (C)2016,JPO&INPIT
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机译：解决的问题：在使用多个秘密分布的秘密计算期间检测伪造。解决方案：秘密计算设备1n输入分布值[a 0 ]，...，[a M-1 ]。随机数生成部12计算分布值[r 0 ]，...，[r J-1 ]。随机化部13计算通过对分布值[a m ]和分布值[a]进行积分而获得的分布值[a m r j ]。 r j ]生成随机分布值 m m ]，[a m r j ]>。保密计算部分14在包括计算对象和函数的同时计算函数值[F（[a 0 ]，...，[a M-1 ]）]]。随机分布值作为校验和C j 中的计算结果。同步部分15等待直到使用所有保密分布的保密计算被终止。正确性证明部分16验证是否通过将分布值[φ j ] [f 0 < / Sub>]，...，分布值[r j ]包含在校验和C j 中的[f μj-1]等于分布值[f 0 r j ]的分布值[ψ j ]，...，[ f μj-1 r j ]包括在校验和C j 中;选定的图纸：图2;版权：（C）2016，JPO＆INPIT
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