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Reminder: Analyzing schemes is off topic on crypto.SE. This question will therefore be likely to be closed soon.

Your scheme can be summarized by:
pick a S-box $\sigma$ and two keys $(k_1, k_2)$ and do:

for i in 0 .. len(message) do
  c[i] = σ(m[i])
  if i & 1:
    σ >>> k2
  else
    σ >>> k1
return c

This is broken by frequency analysis with a sufficiently long message:

• Find $n$ such that $n \times (k_1 + k_2) \equiv 0 \pmod{26}$.
• Remark that every $n \times (k_1 + k_2)$, a character you encrypted with the same key, thus frequency analysis will work.

Worth to be noted:

• this is equivalent to a alphabetic Vigenère Cipher with a long key.
• modern cryptanalysis (CPA) breaks this with a 26 chosen plaintext and 1 chosen ciphertext. See bellow.

Query messages:

AA B C D E F .. Z  

$\alpha\beta$ These will give the $\sigma$ table.
Assuming that AA is encrypted by αβ query a decryption for the ciphertext: $xαβ$ (prepend a random character). Drop the first character decrypted, the following letters will correspond to the shift distances $k_1$ and $k_2$.

In your example, the query would be:
E(AA) = NT
and
D(XNT) = YDH
leading to
D $\implies k_1 = 3$,
H $\implies k_2 = 7$.

Reminder: Analyzing schemes is off topic on crypto.SE. This question will therefore be likely to be closed soon.

Your scheme can be summarized by:
pick a S-box $\sigma$ and two keys $(k_1, k_2)$ and do:

for i in 0 .. len(message) do
  c[i] = σ(m[i])
  if i & 1:
    σ >>> k2
  else
    σ >>> k1
return c

This is broken by frequency analysis with a sufficiently long message:

• Find $n$ such that $n \times (k_1 + k_2) \equiv 0 \pmod{26}$.
• Remark that every $n \times (k_1 + k_2)$, a character you encrypted with the same key, thus frequency analysis will work.

Worth to be noted:

• this is equivalent to a alphabetic Vigenère Cipher with a long key.
• modern cryptanalysis (CPA & CCA) breaks this with a 26 chosen plaintext and 1 chosen ciphertext. See bellow.

Query messages:

AA B C D E F .. Z  

$\alpha\beta$ These will give the $\sigma$ table.
Assuming that AA is encrypted by αβ query a decryption for the ciphertext: $xαβ$ (prepend a random character). Drop the first character decrypted, the following letters will correspond to the shift distances $k_1$ and $k_2$.

In your example, the query would be:
E(AA) = NT
and
D(XNT) = YDH
leading to
D $\implies k_1 = 3$,
H $\implies k_2 = 7$.

Reminder: Analyzing schemes is off topic on crypto.SE. This question will therefore be likely to be closed soon.

Your scheme can be summarized by:
pick a S-box $\sigma$ and two keys $(k_1, k_2)$ and do:

for i in 0 .. len(message) do
  c[i] = σ(m[i])
  if i & 1:
    σ >>> k2
  else
    σ >>> k1
return c

This is broken by frequency analysis with a sufficiently long message:

• Find $n$ such that $n \times (k_1 + k_2) \equiv 0 \pmod{26}$.
• Remark that every $n \times (k_1 + k_2)$, a character you encrypted with the same key, thus frequency analysis will work.

Worth to be noted:

• this is equivalent to a alphabetic Vigenère Cipher with a long key.
• modern cryptanalysis (CPA) breaks this with a 26 chosen plaintext and 1 chosen ciphertext. See bellow.

Query messages:

AA B C D E F .. Z  

$\alpha\beta$ These will give the $\sigma$ table.
Assuming that AA is encrypted by αβ query a decryption for the ciphertext: $xαβ$ (prepend a random character). Drop the first character decrypted, the following letters will correspond to the shift distances $k_1$ and $k_2$. In your example, the query would be  :  XNT, it will return YDH leading to $D = 3, H = 7$.

Reminder: Analyzing schemes is off topic on crypto.SE. This question will therefore be likely to be closed soon.

Your scheme can be summarized by:
pick a S-box $\sigma$ and two keys $(k_1, k_2)$ and do:

for i in 0 .. len(message) do
  c[i] = σ(m[i])
  if i & 1:
    σ >>> k2
  else
    σ >>> k1
return c

This is broken by frequency analysis with a sufficiently long message:

• Find $n$ such that $n \times (k_1 + k_2) \equiv 0 \pmod{26}$.
• Remark that every $n \times (k_1 + k_2)$, a character you encrypted with the same key, thus frequency analysis will work.

Worth to be noted:

• this is equivalent to a alphabetic Vigenère Cipher with a long key.
• modern cryptanalysis (CPA) breaks this with a 26 chosen plaintext and 1 chosen ciphertext. See bellow.

Query messages:

AA B C D E F .. Z  

$\alpha\beta$ These will give the $\sigma$ table.
Assuming that AA is encrypted by αβ query a decryption for the ciphertext: $xαβ$ (prepend a random character). Drop the first character decrypted, the following letters will correspond to the shift distances $k_1$ and $k_2$. 

In your example, the query would be: 
E(AA) = NT
and
D(XNT) = YDH
leading to
D $\implies k_1 = 3$,
H $\implies k_2 = 7$.

Reminder: Analyzing schemes is off topic on crypto.SE. This question will therefore be likely to be closed soon.

Your scheme can be summarized by:
pick a S-box $\sigma$ and two keys $(k_1, k_2)$ and do:

for i in 0 .. len(message) do
  c[i] = σ(m[i])
  if i & 1:
    σ >>> k2
  else
    σ >>> k1
return c

This is broken by frequency analysis with a sufficiently long message:

• Find $n$ such that $n \times (k_1 + k_2) \equiv 0 \pmod{26}$.
• Remark that every $n \times (k_1 + k_2)$, a character you encrypted with the same key, thus frequency analysis will work.

Reminder: Analyzing schemes is off topic on crypto.SE. This question will therefore be likely to be closed soon.

Your scheme can be summarized by:
pick a S-box $\sigma$ and two keys $(k_1, k_2)$ and do:

for i in 0 .. len(message) do
  c[i] = σ(m[i])
  if i & 1:
    σ >>> k2
  else
    σ >>> k1
return c

This is broken by frequency analysis with a sufficiently long message:

• Find $n$ such that $n \times (k_1 + k_2) \equiv 0 \pmod{26}$.
• Remark that every $n \times (k_1 + k_2)$, a character you encrypted with the same key, thus frequency analysis will work.

Worth to be noted:

• this is equivalent to a alphabetic Vigenère Cipher with a long key.
• modern cryptanalysis (CPA) breaks this with a 26 chosen plaintext and 1 chosen ciphertext. See bellow.

Query messages:

AA B C D E F .. Z  

$\alpha\beta$ These will give the $\sigma$ table.
Assuming that AA is encrypted by αβ query a decryption for the ciphertext: $xαβ$ (prepend a random character). Drop the first character decrypted, the following letters will correspond to the shift distances $k_1$ and $k_2$. In your example, the query would be : XNT, it will return YDH leading to $D = 3, H = 7$.