Rebuild the Erasure

Rebuild the Line

Two surviving shard points recover the line in the tiny field.

highlighted = computed this step

Find the slope from kept points

The y change is 1 and the x change is 2 in GF(5).

Δy=1,Δx=2\Delta y=1,\quad \Delta x=2
Rebuild the lineGF(5) erasure recovery rows recomputed from pinned kept shards.Rebuild the line - recover f(1)=0 in GF(5)rowvaluecheckpolynomial2+3x mod 5pinned polynomialshard x=02f(0)shard x=10f(1)shard x=23f(2)shard x=31f(3)missingx=1erasure locationkept(0,2) and (2,3)two surviving shardsdelta y1y2-y0 mod pdelta x2x2-x0 mod pinverse2^-1=3field inverserebuilt slope3delta y times inverserebuilt intercept2y-mxrecovered0f(1)

Divide by multiplying inverse

Division by 2 means multiply by its field inverse 3.

2 inverse 32\text{ inverse }3
Rebuild the lineGF(5) erasure recovery rows recomputed from pinned kept shards.Rebuild the line - recover f(1)=0 in GF(5)rowvaluecheckpolynomial2+3x mod 5pinned polynomialshard x=02f(0)shard x=10f(1)shard x=23f(2)shard x=31f(3)missingx=1erasure locationkept(0,2) and (2,3)two surviving shardsdelta y1y2-y0 mod pdelta x2x2-x0 mod pinverse2^-1=3field inverserebuilt slope3delta y times inverserebuilt intercept2y-mxrecovered0f(1)

The rebuilt slope

The rebuilt slope is 1 times 3, which is 3 modulo 5.

1⋅3≡3(mod5)1\cdot3\equiv3\pmod{5}
Rebuild the lineGF(5) erasure recovery rows recomputed from pinned kept shards.Rebuild the line - recover f(1)=0 in GF(5)rowvaluecheckpolynomial2+3x mod 5pinned polynomialshard x=02f(0)shard x=10f(1)shard x=23f(2)shard x=31f(3)missingx=1erasure locationkept(0,2) and (2,3)two surviving shardsdelta y1y2-y0 mod pdelta x2x2-x0 mod pinverse2^-1=3field inverserebuilt slope3delta y times inverserebuilt intercept2y-mxrecovered0f(1)

The intercept comes back

Using the kept point at x 0 gives rebuilt intercept 2.

b=2b=2
Rebuild the lineGF(5) erasure recovery rows recomputed from pinned kept shards.Rebuild the line - recover f(1)=0 in GF(5)rowvaluecheckpolynomial2+3x mod 5pinned polynomialshard x=02f(0)shard x=10f(1)shard x=23f(2)shard x=31f(3)missingx=1erasure locationkept(0,2) and (2,3)two surviving shardsdelta y1y2-y0 mod pdelta x2x2-x0 mod pinverse2^-1=3field inverserebuilt slope3delta y times inverserebuilt intercept2y-mxrecovered0f(1)