Suyash Bagad
Cryptography Engineer
\(i=0\)
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\(\text{root}\)
\(\text{nodes}\)
\(\text{leaves}\)
\(\text{hash path}\)
\(\text{depth} = 4\)
\(i=0\)
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\(\text{key}\)
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\(h^2_0\)
\(h^1_0\)
\(h^1_1\)
\(i=0\)
\(0\)
\(1\)
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\(\text{key}\)
\(\text{value}\)
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\(h^2_1\)
\(h^2_0\)
\(h^1_0\)
\(h^1_1\)
\(h^2_1\)
\(h^0_6\)
\(2\)
\(\texttt{true}\)
\(h^1_0\)
\(h^0_0\)
\(h^0_1\)
\(h^1_1\)
\(h^0_2\)
\(h^0_3\)
\(1101\)
\(1010\)
\(A\)
\(F\)
\(N\)
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\(i=2\)
\(i=3\)
\(0\)
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\(\text{key}\)
\(\text{value}\)
\(r\)
\(h^3_0\)
\(h^3_1\)
\(h^2_0\)
\(h^3_0\)
\(h^2_1\)
\(5\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(0\)
\(A\)
\(\texttt{true}\)
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\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
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\(3\)
\(A\)
\(F\)
\(N\)
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\(i=1\)
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\(1101\)
\(1010\)
\(r\)
\(h^3_0\)
\(5\)
\(h^2_0\)
\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
\(h^3_0\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
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\(h^3_0\)
\(h^3_1\)
\(A\)
\(F\)
\(N\)
\(i=0\)
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\(i=3\)
\(0\)
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\(1101\)
\(1010\)
\(r\)
\(h^3_0\)
\(5\)
\(h^2_0\)
\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
\(h^3_0\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
\(2\)
\(3\)
\(h^3_0\)
\(h^3_1\)
\(A\)
\(F\)
\(N\)
\(i=0\)
\(i=1\)
\(i=2\)
\(i=3\)
\(0\)
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\(1101\)
\(1010\)
\(r\)
\(h^3_0\)
\(5\)
\(h^2_0\)
\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
\(h^3_0\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
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\(3\)
\(h^3_0\)
\(h^3_1\)
\(\text{Common Height} = 2\)
\(z[2]\)
\(h^2_3\)
\(A\)
\(F\)
\(N\)
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\(i=1\)
\(i=2\)
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\(0\)
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\(1101\)
\(1010\)
\(r\)
\(h^3_0\)
\(5\)
\(h^2_0\)
\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
\(h^3_0\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
\(2\)
\(3\)
\(h^3_0\)
\(h^3_1\)
\(z[2]\)
\(h^2_3\)
\(A\)
\(F\)
\(N\)
\(i=0\)
\(i=1\)
\(i=2\)
\(i=3\)
\(0\)
\(1\)
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\(1101\)
\(1010\)
\(r\)
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\(5\)
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\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
\(h^3_0\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
\(2\)
\(3\)
\(h^3_0\)
\(h^3_1\)
\(z[2]\)
\(h^2_3\)
\(A\)
\(F\)
\(N\)
\(i=0\)
\(i=1\)
\(i=2\)
\(i=3\)
\(0\)
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\(1101\)
\(1010\)
\(r\)
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\(5\)
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\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
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\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
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\(3\)
\(h^3_0\)
\(h^3_1\)
\(z[2]\)
\(h^2_3\)
\(z[1]\)
\(z[1]\)
\(A\)
\(F\)
\(N\)
\(i=0\)
\(i=1\)
\(i=2\)
\(i=3\)
\(0\)
\(1\)
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\(8\)
\(9\)
\(10\)
\(11\)
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\(13\)
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\(15\)
\(1101\)
\(1010\)
\(r\)
\(h^3_0\)
\(5\)
\(h^2_0\)
\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
\(h^3_0\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
\(2\)
\(3\)
\(h^3_0\)
\(h^3_1\)
\(z[2]\)
\(h^2_3\)
\(z[1]\)
\(z[1]\)
\(A\)
\(F\)
\(N\)
\(i=0\)
\(i=1\)
\(i=2\)
\(i=3\)
\(0\)
\(1\)
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\(1101\)
\(1010\)
\(r\)
\(h^3_0\)
\(5\)
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\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
\(h^3_0\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
\(2\)
\(3\)
\(h^3_0\)
\(h^3_1\)
\(z[2]\)
\(h^2_3\)
\(z[1]\)
\(z[1]\)
\(A\)
\(F\)
\(N\)
\(i=0\)
\(i=1\)
\(i=2\)
\(i=3\)
\(0\)
\(1\)
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\(4\)
\(5\)
\(6\)
\(7\)
\(8\)
\(9\)
\(10\)
\(11\)
\(12\)
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\(15\)
\(1101\)
\(1010\)
\(r\)
\(h^3_0\)
\(5\)
\(h^2_0\)
\(\text{key}\)
\(\text{value}\)
\(h^3_1\)
\(h^3_0\)
\(N\)
\(h^3_1\)
\(\texttt{true}\)
\(h^2_1\)
\(0\)
\(A\)
\(\texttt{true}\)
\(1\)
\(F\)
\(\texttt{true}\)
\(h^2_0\)
\(h^2_1\)
\(\vec{h}(10)\)
\(0\)
\(1\)
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\(3\)
\(h^3_0\)
\(h^3_1\)
\(z[2]\)
\(h^2_3\)
\(z[1]\)
\(z[1]\)
\(-1\)
\(-1\)
By Suyash Bagad
How do Merkle trees work?